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RD Sharma Class 12 Exercise 18.20 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.20 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:24 PM IST

The RD Sharma Class 12 Math Solutions have been studied by the whole country for a long time and it has been a huge competition for other publications as well.. RD Sharma Class 12th Exercise 18.20 has 11 questions based on indefinite integrals. The questions in this exercise will be solved only if a student has understood the concept of previous exercise. RD Sharma Solutions This exercise discusses questions based on integrating the function, completing squaring method, and some important formulae used for evaluating the integrals.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.20


Indefinite Integrals Exercise 18.20 Question 1.

Answer: x+\log \left|x^{2}-x\right|+2 \log |x-1|-2 \log |x|+c
Given: \int \frac{x^{2}+x+1}{x^{2}-x} d x
Hint: Using Partial fraction and \int \frac{1}{x} d x
Explanation: Let I=\int \frac{x^{2}+x+1}{x^{2}-x} d x
\frac{x^{2}+x+1}{x^{2}-x}=\frac{x^{2}-x+x+x+1}{x^{2}-x}=\frac{x^{2}-x+2 x+1}{x^{2}-x}
=1+\frac{2 x+1}{x^{2}-x}
\therefore \int \frac{x^{2}+x+1}{x^{2}-x} d x=\int\left(1+\frac{2 x-1+2}{x^{2}-x}\right) d x
\begin{aligned} &=\int 1 d x+\int \frac{2 x-1}{x^{2}-x} d x+2 \int \frac{1}{x^{2}-x} d x\\ &=x+\log \left|x^{2}-x\right|+2 I_{1} \quad \ldots \ldots \end{aligned}(1)
where, I_{1}=\int \frac{1}{x^{2}-x} d x
Now,\int \frac{1}{x^{2}-x} d x=\int \frac{1}{x(x-1)} d x
\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}
Multiplying by x(x-1)
1=A(x-1)+B(x)
Putting
\begin{aligned} &x=1 \\ &1=A(1-1)+B(1) \Rightarrow B=1 \end{aligned}
Putting
\begin{aligned} &x=0 \\ &1=A(0-1)+B(0) \Rightarrow A=-1 \end{aligned}
\begin{aligned} &\therefore \frac{1}{x(x-1)}=\frac{-1}{x}+\frac{1}{x-1} \\ &\therefore \int \frac{1}{x(x-1)} d x=-\int \frac{1}{x} d x+\int \frac{1}{x-1} d x \\ &\therefore I_{1}=-\log |x|+\log |x-1| \end{aligned}
Put in (1)
\begin{aligned} &I=x+\log \left|x^{2}-x\right|+2[-\log |x|+\log |x-1|] \\ &=x+\log \left|x^{2}-x\right|-2 \log |x|+2 \log |x-1|+c \end{aligned}

Indefinite Integrals Exercise 18.20 Question 2.

Given: x+\log \left|\frac{x-2}{x+3}\right|+c
Hint: Using Partial Fraction
Explanation:
I=\int \frac{x^{2}+x-1}{x^{2}+x-6} d x
\frac{x^{2}+x-1}{x^{2}+x-6}=\frac{x^{2}+x-6+5}{x^{2}+x-6}=1+\frac{5}{x^{2}+x-6}
=1+\frac{5}{(x-2)(x+3)}
\left[\begin{array}{l} x^{2}+x-6=x^{2}+3 x-2 x-6 \\ =x(x+3)-2(x+3) \\ =(x-2)(x+3) \end{array}\right]
\because \int \frac{x^{2}+x-1}{x^{2}+x-6} d x=\int 1 d x+5 \int \frac{1}{(x-2)(x+3)} d x
\because I=x+5 I_{1} .......................(1)
Where I_{1}=\int \frac{1}{(x-2)(x+3)} d x
\frac{1}{(x-2)(x+3)}=\frac{A}{(x-2)}+\frac{B}{(x+3)}
Multiplying by (x-2)(x+3)
1=A(x+3)+B(x-2)
Put x=3
1=A(0)+B(-5) \Rightarrow B=\frac{-1}{5}
Put x=2
1=A(5)+B(0) \Rightarrow A=\frac{1}{5}
\therefore \frac{1}{(x-2)(x+3)}=\frac{\frac{1}{5}}{(x-2)}+\frac{-\frac{1}{5}}{(x+3)}
\therefore \int \frac{1}{(x-2)(x+3)} d x=\frac{1}{5} \int \frac{1}{x-2} d x+\left(\frac{-1}{5}\right) \int \frac{1}{x+3} d x
\begin{aligned} &=\frac{1}{5} \log |x-2|-\frac{1}{5} \log |x+3|+c \\ &=\frac{1}{5} \log \left|\frac{x-2}{x+3}\right|+c \end{aligned}
Put in (1)
\begin{aligned} &I=x+5\left(\frac{1}{5} \log \left|\frac{x-2}{x+3}\right|\right)+c \\ &I=x+\log \left|\frac{x-2}{x+3}\right|+c \end{aligned}

Indefinite Integrals Exercise 18.20 Question 3.

Answer: \frac{x}{2}+\log |x|-\frac{3}{4} \log |2 x-1|+c
Given: \int \frac{\left(1-x^{2}\right)}{x(1-2 x)} d x
Hint: Using Partial fraction
Explanation:
LetI=\int \frac{\left(1-x^{2}\right)}{x(1-2 x)} d x=\int \frac{1-x^{2}}{x-2 x^{2}} d x
I=\int \frac{x^{2}-1}{2 x^{2}-x} d x
\frac{x^{2}-1}{2 x^{2}-x}=\frac{1}{2}+\frac{\frac{1}{2} x-1}{2 x^{2}-x}
\therefore \int \frac{x^{2}-1}{2 x^{2}-x} d x=\int \frac{1}{2} d x+\int \frac{\frac{x}{2}-1}{2 x^{2}-x} d x
=\frac{1}{2} x+I_{1} ............(1) Where I_{1}=\int \frac{\frac{x}{2}-1}{2 x^{2}-x} d x
\frac{\frac{x}{2}-1}{2 x^{2}-x}=\frac{\frac{x}{2}-1}{x(2 x-1)}=\frac{A}{x}+\frac{B}{2 x-1}
Multiplying by x(2 x-1)
\frac{x}{2}-1=A(2 x-1)+B(x)
Putting x = 0
-1=A(-1)+B(0) \Rightarrow A=1
Putting x = 2
\begin{aligned} &1-1=1(4-1)+B(2) \\ &0=3+2 B \\ &B=\frac{-3}{2} \end{aligned}
\begin{aligned} &\therefore \frac{\frac{x}{2}-1}{2 x^{2}-x}=\frac{1}{x}-\frac{3}{2(2 x-1)} \\ &\therefore I_{1}=\int \frac{1}{x} d x-\frac{3}{2} \int \frac{2}{2(2 x-1)} d x \\ &=\int \frac{1}{x} d x-\frac{3}{4} \int \frac{2}{(2 x-1)} d x \\ &=\log |x|-\frac{3}{4} \log |2 x-1| \end{aligned}
Put in (1)
\frac{1}{2} x+\log |x|-\frac{3}{4} \log |2 x-1|+c

Indefinite Integrals Exercise 18.20 Question 4.

Answer: 10 \log |x-3|-5 \log |x-2|+c
Given: \int \frac{x^{2}+1}{x^{2}-5 x+6} d x
Hint: Using partial fraction and \int \frac{1}{x} d x
Explanation:
\begin{aligned} &\text { Let }\\ &I=\int \frac{x^{2}+1}{x^{2}-5 x+6} d x \end{aligned}
\frac{x^{2}+1}{x^{2}-5 x+6}=\frac{x^{2}+1}{(x-3)(x-2)} \quad\left[\begin{array}{l} x^{2}-5 x+6=x^{2}-3 x-2 x+6 \\ =x(x-3)-2(x-3) \\ =(x-3)(x-2) \end{array}\right]
\frac{x^{2}+1}{x^{2}-5 x+6}=\frac{A}{(x-3)}+\frac{B}{(x-2)}
Multiplying by \left ( x-3 \right ) and \left ( x-2\right )
x^{2}+1=A(x-2)+B(x-3)
Putting x = 2
4+1=A(0)+B(-1) \Rightarrow B=-5
Putting x = 3
9+1=A(3-2)+B(0) \Rightarrow A=10
\therefore \int \frac{x^{2}+1}{x^{2}-5 x+6} d x=10 \int \frac{1}{x-3} d x-5 \int \frac{1}{x-2} d x
=10 \log |x-3|-5 \log |x-2|+c

Indefinite Integrals Exercise 18.20 Question 5.

Answer: -\frac{25}{3} \log |x+5|+\frac{4}{3} \log |x+2|+c
Given: \int \frac{x^{2}}{x^{2}+7 x+10} d x
Hint: Using Partial Fraction and \int \frac{1}{x} d x
Explanation:
Let I=\int \frac{x^{2}}{x^{2}+7 x+10} d x
\frac{x^{2}}{x^{2}+7 x+10}=\frac{x^{2}}{x^{2}+5 x+2 x+10}=\frac{x^{2}}{x(x+5)+2(x+5)}=\frac{x^{2}}{(x+5)(x+2)}
\frac{x^{2}}{(x+5)(x+2)}=\frac{A}{x+5}+\frac{B}{x+2}
Multiply by (x+5)(x+2)
x^{2}=A(x+2)+B(x+5)
Put x = -2
4=A(0)+B(3) \Rightarrow B=\frac{4}{3}
Put x = -5
25=A(-3)+B(0) \Rightarrow A=-\frac{25}{3}
\frac{x^{2}}{(x+5)(x+2)}=\frac{\frac{-25}{3}}{x+5}+\frac{\frac{4}{3}}{x+2}
\therefore \int \frac{x^{2}}{(x+5)(x+2)} d x=\frac{-25}{3} \int \frac{1}{x+5} d x+\frac{4}{3} \int \frac{1}{x+2} d x
=\frac{-25}{3} \log |x+5|+\frac{4}{3} \log |x+2|+c

Indefinite Integrals Exercise 18.20 Question 6.

Answer: x+\log \left|x^{2}-x+1\right|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c
Given: \int \frac{x^{2}+x+1}{x^{2}-x+1} d x
Hint: Using Partial Fraction
Explanation:
Let
I=\int \frac{x^{2}+x+1}{x^{2}-x+1} d x
\frac{x^{2}+x+1}{x^{2}-x+1}=\frac{x^{2}-x+x+x+1}{x^{2}-x+1}=\frac{x^{2}-x+1}{x^{2}-x+1}+\frac{2 x}{x^{2}-x+1}
=1+\frac{2 x-1}{x^{2}-x+1}+\frac{1}{x^{2}-x+1}
\int \frac{x^{2}+x+1}{x^{2}-x+1} d x=\int 1 d x+\int \frac{2 x-1}{x^{2}-x+1} d x+\int \frac{1}{x^{2}-x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x
=\int 1 d x+\int \frac{2 x-1}{x^{2}-x+1} d x+\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x
=x+\log \left|x^{2}-x+1\right|+\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c
=x+\log \left|x^{2}-x+1\right|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c

Indefinite Integrals Exercise 18.20 Question 7.

Answer: x-2 \log \left|x^{2}+x+2\right|+3 \tan ^{-1}(x+1)+c
Given: \int \frac{(x-1)^{2}}{x^{2}+2 x+2} d x
Hint: \text { Using } \int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x
Explanation: Let
I=\int \frac{(x-1)^{2}}{x^{2}+2 x+2} d x=\int \frac{x^{2}-2 x+1}{x^{2}+2 x+2} d x
=1-\frac{(4 x+1)}{x^{2}+2 x+2}
=1-2\left(\frac{2 x+\frac{1}{2}}{x^{2}+2 x+2}\right)
=1-2\left(\frac{2 x+2-2+\frac{1}{2}}{x^{2}+2 x+2}\right)
=1-2\left(\frac{2 x+2-2+\frac{1}{2}}{x^{2}+2 x+2}\right)
=1-2\left(\frac{2 x+2}{x^{2}+2 x+2}\right)-2\left(\frac{-\frac{3}{2}}{x^{2}+2 x+(1)^{2}-(1)^{2}+2}\right)
=1-2\left(\frac{2 x+2}{x^{2}+2 x+2}\right)+3\left(\frac{1}{(x+1)^{2}+1}\right)
\therefore \int \frac{(x-1)^{2}}{x^{2}+2 x+2} d x=\int 1 d x-2 \int \frac{2 x+2}{x^{2}+2 x+2} d x+3 \int \frac{1}{(x+1)^{2}+1} d x
=x-2 \log \left|x^{2}+2 x+2\right|+3 \tan ^{-1}(x+1)+c


Indefinite Integrals Exercise 18.20 Question 8.

Answer: \frac{x^{2}}{2}+2 x+\frac{3}{2} \log \left|x^{2}-x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c
Given: \int \frac{x^{3}+x^{2}+2 x+1}{x^{2}-x+1} d x
Hint: Useing \int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x
Explanation:
I=\int \frac{x^{3}+x^{2}+2 x+1}{x^{2}-x+1} d x
\frac{x^{3}+x^{2}+2 x+1}{x^{2}-x+1}=(x+2)+\left(\frac{3 x-1}{x^{2}-x+1}\right)
\therefore \int \frac{x^{3}+x^{2}+2 x+1}{x^{2}-2 x+1} d x=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-\frac{2}{3}}{x^{2}-x+1} d x
=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-1+1-\frac{2}{3}}{x^{2}-x+1} d x
=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-1}{x^{2}-x+1} d x+\frac{3}{2} \times \frac{1}{3} \int \frac{1}{x^{2}-x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x
=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-1}{x^{2}-x+1} d x+\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x
=\frac{x^{2}}{2}+2 x+\frac{3}{2} \log \left|x^{2}-x+1\right|+\frac{1}{2} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c
=\frac{x^{2}}{2}+2 x+\frac{3}{2} \log \left|x^{2}-x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c


Indefinite Integrals Exercise 18.20 Question 9.

Answer: \frac{x^{5}}{5}-\frac{4 x^{3}}{3}+20 x-40 \tan ^{-1} \frac{x}{2}+c
Hint: Using \int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x
Explanation: Let
I=\int \frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4} d x
\frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4}=\frac{\left(x^{2}+4-4\right)\left(x^{4}+4\right)}{x^{2}+4}=\frac{\left[\left(x^{2}+4\right)-4\right]\left(x^{4}+4\right)}{x^{2}+4}
=\frac{\left(x^{2}+4\right)\left(x^{4}+4\right)}{x^{2}+4}-\frac{4\left(x^{4}+4\right)}{x^{2}+4}
\therefore \int \frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4} d x=\int\left(x^{4}+4\right) d x-4 \int \frac{x^{4}}{x^{2}+4} d x-16 \int \frac{1}{x^{2}+4} d x
=\int\left(x^{4}+4\right) d x-4 \int\left(x^{2}-\frac{4 x^{2}}{x^{2}+4}\right) d x-16 \int \frac{1}{x^{2}+4} d x
\left[\because \frac{x^{4}}{x^{2}+4}=x^{2}-\frac{4 x^{2}}{x^{2}+4}\right]
=\int\left(x^{4}+4\right) d x-4 \int x^{2} d x+16 \int \frac{x^{2}+4-4}{x^{2}+4} d x-16 \int \frac{1}{x^{2}+4} d x
=\int\left(x^{4}+4\right) d x-4 \int x^{2} d x+16 \int 1 d x-64 \int \frac{d x}{x^{2}+4}-16 \int \frac{d x}{x^{2}+4}
=\frac{x^{5}}{5}+4 x-\frac{4 x^{3}}{3}+16 x-\frac{64}{2} \tan ^{-1} \frac{x}{2}-\frac{16}{2} \tan ^{-1} \frac{x}{2}+c
=\frac{x^{5}}{5}+20 x-\frac{4 x^{3}}{3}-40 \tan ^{-1} \frac{x}{2}+c
=\frac{x^{5}}{5}-\frac{4 x^{3}}{3}+20 x-40 \tan ^{-1} \frac{x}{2}+c

Indefinite Integrals Exercise 18.20 Question 10.

Answer: x-3 \log \left|x^{2}+6 x+12\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c
Given: \int \frac{x^{2}}{x^{2}+6 x+12} d x
Hint: Using\int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x
Explanation: Let
I=\int \frac{x^{2}}{x^{2}+6 x+12} d x
\begin{aligned} &=\int\left(1-\left(\frac{6 x+12}{x^{2}+6 x+12}\right)\right) d x \\ &=\int 1 d x-6 \int \frac{x+2}{x^{2}+6 x+12} d x \\ &=\int 1 d x-\frac{6}{2} \int \frac{2 x+4+2-2}{x^{2}+6 x+12} d x \end{aligned}
\begin{aligned} &=\int 1 d x-3 \int \frac{2 x+6}{x^{2}+6 x+12} d x+6 \int \frac{1}{x^{2}+6 x+(3)^{2}-(3)^{2}+12} d x \\ &=\int 1 d x-3 \int \frac{2 x+6}{x^{2}+6 x+12}+6 \int \frac{1}{(x+3)^{2}+(\sqrt{3})^{2}} d x \end{aligned}
\begin{aligned} &=x-3 \log \left|x^{2}+6 x+12\right|+6 \times \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c \\ &=x-3 \log \left|x^{2}+6 x+12\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c \end{aligned}

Indefinite Integrals Exercise 18.20 Question 11.

Answer:x+\log \left|\frac{x}{x+1}\right|+c
Given: \int \frac{x^{3}+1}{x^{3}-x} d x
Hint: using Partial Fraction and \int \frac{1}{x} d x
Explanation: Let
I=\int \frac{x^{3}+1}{x^{3}-x} d x
\frac{x^{3}+1}{x^{3}-x}=\left(1+\frac{x+1}{x^{3}-x}\right)=1+\frac{x+1}{x\left(x^{2}-1\right)}
=1+\frac{x+1}{x(x-1)(x+1)}=1+\frac{1}{x(x-1)}
\therefore \int \frac{x^{3}+1}{x^{3}-x} d x=\int 1 d x+\int \frac{1}{x(x+1)} d x
=x+I_{1} ...................(1) Where I_{1}=\int \frac{1}{x\left ( x+1 \right )}dx
\frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{(x+1)}

Multiplying by x\left ( x+1 \right )

1=A\left ( x+1 \right )+B\left ( x+1 \right )
Putting x = -1
1=A\left ( 0+1 \right )+B\left ( 0 \right )\Rightarrow A=1
\begin{aligned} &\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{(x+1)} \\ &\therefore \int \frac{1}{x(x+1)} d x=\int \frac{1}{x} d x-\int \frac{1}{x+1} d x \end{aligned}
=\log |x|-\log |x+1|
Put in (1)
\begin{aligned} I &=x+\log |x|-\log |x+1|+c \\ &=x+\log \left|\frac{x}{x+1}\right|+c \end{aligned}


Indefinite Integrals Exercise 18.12 Question 9

Answer: - -\frac{1}{6} \cos ^{6} x+\frac{1}{8} \cos ^{8} x+C
Hint: - Use substitution method to solve this integral.
Given: -\int \sin ^{3} x \cdot \cos ^{5} x d x
Solution: - Let I=\int \sin ^{3} x \cdot \cos ^{5} x d x
Substitute \cos x=t \Rightarrow-\sin x d x=d t then
\begin{aligned} I &=\int\sin ^{3} x t^{5} \cdot \frac{d t}{-\sin x}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left [ \because \cos x=t \right ] \\ &=-\int \sin ^{2} x t^{5} \cdot d t=-\int\left(t-\cos ^{2} x\right) t^{5} d t \quad \quad \quad \quad \quad \quad \quad \left [ \because \sin ^{2}x+\cos ^{2 }x=1 \right ]\\ &=-\int\left(1-t^{2}\right)\left(t^{5}\right) d t=-\int\left(t^{5}-t^{5} t^{2}\right) d t \end{aligned}
\begin{aligned} &=-\int\left(t^{5}-t^{7}\right) d t=-\int t^{5} d t+\int t^{7} d t \\ &=-\frac{t^{5+1}}{5+1}+\frac{t^{7+1}}{7+1}+C \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=-\frac{t^{6}}{6}+\frac{t^{8}}{8}+c=-\frac{\cos ^{6} x}{6}+\frac{\cos ^{8} x}{8}+C \quad\quad\quad\quad\quad\quad\quad[\because \cos x=t] \end{aligned}

Indefinite Integrals Exercise 18.12 Question 11

Answer: --\frac{1}{2} \tan ^{-2} x+3 \log |\tan x|+\frac{3}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C
Hint: - Use substitution method to solve this integral.
Given: -\int \frac{1}{\sin ^{3} x \cdot \cos ^{5} x} d x
The exponent of denominator is 3+5=8. To solve this type of integral, we have to divide both numerator and denominator by \cos ^{8 }x then
\begin{aligned} &I=\int \frac{\frac{1}{\cos ^{8} x}}{\frac{\sin ^{3} x \cdot \cos ^{5} x}{\cos ^{8} x}} d x=\int \frac{\sec ^{8} x}{\frac{\sin ^{3} x}{\cos ^{3} x}} d x \\ &=\int \frac{\sec ^{8} x}{\tan ^{3} x} d x=\int \frac{\sec ^{6} x \cdot \sec ^{2} x}{\tan ^{3} x} d x=\int \frac{\left(\sec ^{2} x\right)^{3} \sec ^{2} x}{\tan ^{3} x} \end{aligned}
\begin{aligned} &=\int \frac{\left(1+\tan ^{2} x\right)^{3} \sec ^{2} x}{\tan ^{2} x} d x \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\because 1+\tan ^{2} x=\sec ^{2} x\right] \\ &\Rightarrow I=\int \frac{\left(1+\tan ^{6} x+3 \tan ^{2} x+3 \tan ^{4} x\right) \sec ^{2} x}{\tan ^{3} x} d x \end{aligned}
Substitute \tan x=t \Rightarrow \sec ^{3} x d x=d t then
\begin{aligned} I &=\int \frac{\left(1+t^{6}+3 t^{2}+3 t^{4}\right)}{t^{3}} d t=\int\left(\frac{1}{t^{3}}+\frac{t^{6}}{t^{3}}+\frac{3 t^{2}}{t^{3}}+\frac{3 t^{4}}{t^{3}}\right) d t \\ &=\int\left(t^{-3}+t^{6-3}+\frac{3}{t}+3 t\right) d t=\int\left(t^{-3}+t^{3}+\frac{3}{t}+3 t\right) d t \\ &=\int t^{-3} d t+\int t^{3} d t+3 \int_{t}^{1} \frac{1}{t} d t+3 \int t d t \end{aligned}
=\frac{t^{-3+1}}{-3+1}+\frac{t^{3+1}}{3+1}+3 \log |t|+\frac{3 t^{1+1}}{1+1}+C \quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \frac{1}{t} d t=\log |t|+C\right]
\begin{aligned} &=\frac{t^{-2}}{-2}+\frac{t^{4}}{4}+3 \log |t|+\frac{3 t^{2}}{2}+C \\ &=-\frac{1}{2} t^{-2}+3 \log |t|+\frac{3 t^{2}}{2}+\frac{1}{4} t^{4}+C \\ &=-\frac{1}{2 \tan ^{2} x}+3 \log |\tan x|+\frac{3}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C \quad\quad\quad\quad[\because t=\tan x] \end{aligned}


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These solutions are the best source if a student wants to learn and understand the concept fast. The students will be able to revise at a fast pace with RD Sharma Class 12 Solutions Chapter 18 ex 18.20.

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RD Sharma Class 12 Solutions Indefinite Integrals Ex. 18.20 helps a student to learn from NCERT books because these solutions contain NCERT type questions too.

  1. Various methods to solve a single question:

RD Sharma Class 12th Exercise 18.20 helps students find alternative methods to solve a single question. Team of subject experts take utmost care of the type of questions asked and provide answers in such a way that a student can choose which is suitable for him.

  1. Exceptional performance in exams

These solutions help a student score high grades in exams. RD Sharma Class 12 Solutions Ex. 18.20 has all the questions that fulfil all the requirements of exams.

  1. Free of Cost

There is a misconception that the authorized set of RD Sharma books cost a lot. But the fact is that, the students can download these books from the career 360 website without paying even a single penny.

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. What are the bounds of the indefinite integrals?

There are no bounds for indefinite integrals.

2. Which maths textbook is the best for Class 12?

RD Sharma Textbook is the most prestigious book for the Class 12 exam. The CBSE curriculum is strictly followed in RD Sharma's books. It serves as a foundation for preparing for board exams because it contains question papers based on the CBSE syllabus for the subject.

3. Are RD Sharma Class 12th Exercise 18.20 solutions free?

Yes, RD Sharma Class 12th Exercise 18.20 solutions are present in the Career 360 website for free. 


4. What are the benefits of RD Sharma solutions?

The Advantages of RD Sharma Solutions for Class 12 Maths:

  • Career360 expert solutions prepare these solutions with a focus on accuracy.

  • The solutions to each exercise are easily accessible from the chapters, which are available in pdf format.

  • Students can use these books to assess their knowledge gaps and study accordingly.

5. What is the indefinite integral of 0?

The integral of 0 is C because the derivative of C (or any constant) is zero. Therefore, ∫0 dx = C.

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