What are the bounds of the indefinite integrals?

There are no bounds for indefinite integrals.

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.

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.

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.

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.

RD Sharma Class 12 Exercise 18.20 Indefinite Integrals Solutions Maths

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:
Let $I=\int \frac{\left(1-x^{2}\right)}{x(1-2 x)} d x=\int \frac{1-x^{2}}{x-2 x^{2}} d x$
$I=\int \frac{x^{2}-1}{2 x^{2}-x} d x$
$\frac{x^{2}-1}{2 x^{2}-x}=\frac{1}{2}+\frac{\frac{1}{2} x-1}{2 x^{2}-x}$
$\therefore \int \frac{x^{2}-1}{2 x^{2}-x} d x=\int \frac{1}{2} d x+\int \frac{\frac{x}{2}-1}{2 x^{2}-x} d x$
$=\frac{1}{2} x+I_{1}$ ............(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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