RD Sharma Class 12 Exercise 18.13 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.13 Indefinite Integrals Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.13
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.13

Indefinite Integrals Excercise 18.13 Question 2

Answer:

\frac{1}{8 a^{2}} \frac{x^{8}}{{(a^{2} - x^{2})}^{4}} + c

Hint: Use substitution method to solve this integral.
Given:

\int \frac{x^{7}}{{(a^{2} - x^{2})}^{5}} d x

Solution:
Let

I = \int \frac{x^{7}}{{(a^{2} - x^{2})}^{5}} d x

Put

x = a \sin θ \Rightarrow d x = a \cos θ d θ

…(i) (Differentiating w.r.t to θ)

\begin{aligned} Then I = \int \frac{a^{7} \sin^{7} θ}{{(a^{2} - a^{2} \sin^{2} θ)}^{5}} \cdot a \cos θ d θ \\ = \int \frac{a^{8} \sin^{7} θ \cos θ}{{a^{2} (1 - \sin^{2} θ)}^{5}} d θ = \int \frac{a^{8} \sin^{7} θ \cos θ}{{a^{2 \times 5} (\cos^{2} θ)}}^{5}} d θ \\ = \int \frac{a^{8} \sin^{7} θ \cos θ}{a^{10} \cos^{10} θ} d θ = \frac{1}{a^{2}} \int \frac{\sin^{7} θ}{\cos^{9} θ} d θ \end{aligned}

\begin{aligned} = \frac{1}{a^{2}} \int \frac{\sin^{7} θ}{\cos^{7} θ} \cdot \frac{1}{\cos^{2} θ} d θ \\ = \frac{1}{a^{2}} \int \tan^{7} θ \cdot \sec^{2} θ d θ [∵ \frac{\sin θ}{\cos θ} = \tan θ and \frac{1}{\cos θ} = \sec θ] \end{aligned}

Again put

\tan θ = t \Rightarrow \sec 2 θ d θ = d t

(on differentiating w.r.t to

θ

)

\begin{aligned} Then I = \frac{1}{a^{2}} \int t^{7} d t = \frac{1}{a^{2}} \cdot \frac{t^{7 + 1}}{7 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ I = \frac{1}{a^{2}} \cdot \frac{t^{8}}{8} + c = \frac{1}{8 a^{2}} \cdot t^{8} + c \\ I = \frac{1}{8 a^{2}} \cdot t^{8} + c [∵ t = \tan θ] \end{aligned}

From (i)

x = a \sin θ

\Rightarrow \sin θ = \frac{x}{a}

and we know that

\sin^{2} θ + \cos^{2} θ = 1

\begin{aligned} \Rightarrow \cos^{2} θ = 1 - \sin^{2} θ \\ \Rightarrow \cos θ = \sqrt{1 - \sin^{2} θ} = \sqrt{1 - {(\frac{x}{a})}^{2}} \\ \Rightarrow \cos θ = \sqrt{1 - \frac{x^{2}}{a^{2}}} = \sqrt{\frac{a^{2} - x^{2}}{a^{2}}} = \frac{1}{a} \sqrt{a^{2} - x^{2}} \end{aligned}

\begin{aligned} ∴ \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{x}{a}}{\frac{1}{a} \sqrt{a^{2} - x^{2}}} \\ \Rightarrow \tan θ = \frac{x}{a} \cdot \frac{a}{\sqrt{a^{2} - x^{2}}} = \frac{x}{\sqrt{a^{2} - x^{2}}} \end{aligned}

∴ \tan θ = \frac{x}{\sqrt{a^{2} - x^{2}}}

…(iii)

Put the value of tanθ from (iii) in (ii) then

\begin{aligned} I = \frac{1}{8 a^{2}} {(\frac{x}{\sqrt{a^{2} - x^{2}}})}^{8} + c \\ = \frac{1}{8 a^{2}} \frac{x^{8}}{{(a^{2} - x^{2})}^{\frac{8}{2}}} + c \\ = \frac{1}{8 a^{2}} \frac{x^{8}}{{(a^{2} - x^{2})}^{4}} + c \end{aligned}

Indefinite Integrals Excercise 18.13 Question 3

Answer:

\frac{x^{2}}{2} + c

Hint: Use substitution method to solve this integral
Given:

\int \cos {2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}}} d x

Put

x = \cos 2 θ

…(i)

\begin{aligned} Then I = \int \cos {2 \cot^{- 1} \sqrt{\frac{1 + \cos 2 θ}{1 - \cos 2 θ}}} d x \\ = \int \cos {2 \cot^{- 1} \sqrt{\frac{2 \cos^{2} θ}{2 \sin^{2} θ}}} d x {\begin{matrix} ∵ 1 + \cos 2 A = 2 \cos^{2} A \\ 1 - \cos 2 A = 2 \sin^{2} A \end{matrix}} \\ = \int \cos {2 \cot^{- 1} \sqrt{{(\frac{\cos θ}{\sin θ})}^{2}}} d x \\ = \int \cos {2 \cot^{- 1} \sqrt{(\cot θ)^{2}}} d x [∵ \frac{\cos A}{\sin A} = \cot A] \\ = \int \cos {2 \cot^{- 1} \cdot \cot θ} d x \end{aligned}

\begin{aligned} = \int \cos {2 θ} d x [∵ \cot (\cot^{- 1} x) = x] \\ = \int \cos {\cos^{- 1} (x)} d x \\ = \int x^{1} d x [\begin{array}{c} ∵ x = \cos 2 θ \\ 2 θ = \cos^{- 1} (x) \end{array}] \\ = \frac{x^{1 + 1}}{1 + 1} + c [∵ \cos (\cos^{- 1} x) = x] \\ = \frac{x^{2}}{2} + c \end{aligned}

Indefinite Integrals Excercise 18.13 Question 4

Answer: $\frac{-1}{3} \frac{\left(1+x^{2}\right)^{\frac{3}{2}}}{x^{3}}+c $
Hint: Use substitution method to solve this integral
Given:

\int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x

Solution:
Let

I = \int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x

Putting

x = \tan θ

… (i)

d x = \sec 2 θ d θ

Then

\begin{aligned} I = \int \frac{\sqrt{1 + \tan^{2} θ}}{\tan^{4} θ} \sec^{2} θ d θ \\ = \int \frac{\sqrt{\sec^{2} θ}}{\tan^{4} θ} \cdot \sec^{2} θ d θ [\begin{matrix} ∵ \sec^{2} θ - \tan^{2} θ = 1 \\ 1 + \tan^{2} θ = \sec^{2} θ \end{matrix}] \\ = \int \frac{\sec θ \cdot \sec^{2} θ}{\tan^{4} θ} d θ = \int \frac{\sec^{3} θ}{\tan^{4} θ} d θ \\ = \int \frac{1}{\cos^{3} θ \cdot \frac{\sin^{4} θ}{\cos^{4} θ}} d θ = \int \frac{\cos^{4} θ}{\sin^{4} θ \cdot \cos^{3} θ} d θ \\ = \int \frac{\cos θ}{\sin^{4} θ} d θ \end{aligned}

Again putting

\sin θ = t

\begin{aligned} \Rightarrow \cos θ d θ = d t then \\ I = \int \frac{1}{t^{4}} \cdot d t \\ = \int t^{- 4} d t = \frac{t^{- 4 + 1}}{- 4 + 1} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ = \frac{t^{- 3}}{- 3} + c \\ = \frac{- 1}{3} t^{- 3} + c \\ = \frac{- 1}{3 t^{3}} + c u. (ii) \end{aligned}

Also from (i),

x = \tan θ

\begin{aligned} x^{2} = \tan^{2} θ [Squaring both sides] \\ \Rightarrow x^{2} = \frac{\sin^{2} θ}{\cos^{2} θ} [\tan θ = \frac{\sin θ}{\cos θ}] \\ \Rightarrow x^{2} \cos^{2} θ = \sin^{2} θ \\ \Rightarrow x^{2} (1 - \sin^{2} θ) = \sin^{2} θ [\begin{matrix} \sin^{2} θ + \cos^{2} θ = 1 \\ \Rightarrow \sin^{2} θ = 1 - \cos^{2} θ \end{matrix}] \end{aligned}

\begin{aligned} \Rightarrow x^{2} - x^{2} \sin^{2} θ = \sin^{2} θ \\ \Rightarrow x^{2} = \sin^{2} θ + x^{2} \sin^{2} θ = \sin^{2} θ (1 + x^{2}) \\ \Rightarrow \sin^{2} θ = \frac{x^{2}}{1 + x^{2}} \\ \Rightarrow \sin θ = \sqrt{\frac{x^{2}}{1 + x^{2}}} \\ \Rightarrow \sin θ = \frac{x}{\sqrt{1 + x^{2}}} \end{aligned}

Putting the value of sinθ from (iii) in (ii), we get

I = \frac{- 1}{3} \frac{1}{{(\sin θ)}^{3}} + c

\begin{aligned} = \frac{- 1}{3} \frac{1}{{(\frac{x}{\sqrt{1 + x^{2}}})}^{3}} + c \\ = \frac{- 1}{3} \frac{1}{\frac{x^{3}}{{(\sqrt{1 + x^{2}})}^{3}}} + c \\ = - \frac{1}{3} \frac{{(1 + x^{2})}^{\frac{3}{2}}}{x^{3}} + c \end{aligned}

Indefinite Integrals Excercise 18.13 Question 1

Answer:

\frac{x}{\sqrt{a^{2} - x^{2}}} - \sin^{- 1} (\frac{x}{a}) + c

Hint: Use substitution method to solve this integral.
Given:

\int \frac{x^{2}}{{(a^{2} - x^{2})}^{\frac{3}{2}}} d x

Solution:
Let

I = \int \frac{x^{2}}{{(a^{2} - x^{2})}^{\frac{3}{2}}} d x

Put

x = a \sin u

…(i)

d x = a \cos u d u

(on differentiating w.r.t to u)
Then

\begin{aligned} I = \int \frac{a^{2} \sin^{2} u}{{(a^{2} - a^{2} \sin^{2} u)}^{\frac{3}{2}}} \cdot a \cos u d u = \int \frac{a^{3} \sin^{2} u \cdot \cos u}{{(a^{2} (1 - \sin^{2} u)}}^{\frac{3}{2}}} d u \\ = \int \frac{a^{3} \sin^{2} u \cdot \cos u}{{(a^{2})}^{\frac{3}{2}} {(1 - \sin^{2} u)}^{\frac{3}{2}}} d u = \int \frac{a^{3} \sin^{2} u \cdot \cos u}{(a^{3}) {(\cos^{2} u)}^{\frac{3}{2}}} d u \\ = \int \frac{a^{3} \sin^{2} u \cdot \cos u}{(a^{3}) {(\cos^{2} u)}^{\frac{3}{2}}} d u \\ = \int \frac{\sin^{2} u \cdot \cos u}{(\cos u)^{2 \times \frac{3}{2}}} d u \end{aligned}

\begin{aligned} = \int \frac{\sin^{2} u \cdot \cos u}{(\cos u)^{3}} d u \\ = \int \frac{\sin^{2} u}{(\cos u)^{2}} d u = \int {(\frac{\sin u}{\cos u})}^{2} d u \\ = \int (\tan u)^{2} d u = \int \tan^{2} u d u [∵ \frac{\sin θ}{\cos θ} = \tan θ] \end{aligned}

\begin{aligned} = \int (\sec^{2} u - 1) d u = \int \sec^{2} u d u - \int 1 d u \\ [∵ 1 + \tan^{2} θ = \sec^{2} θ \Rightarrow \tan^{2} θ = \sec^{2} θ - 1] \\ I = \tan u - u + c ... (ii) [∵ \int 1 d x = x + c] \end{aligned}

From (i)

x = a \sin u

\Rightarrow \sin u = \frac{x}{a} \Rightarrow u = \sin^{- 1} (\frac{x}{a})

…(iii)
And we know that

1 = \sin^{2} u + \cos^{2} u

then

\begin{aligned} \Rightarrow \cos^{2} u = 1 - \sin^{2} u \\ \Rightarrow \cos u = \sqrt{1 - \sin^{2} u} = \sqrt{1 - {(\frac{x}{a})}^{2}} \end{aligned}

Now

\begin{aligned} \tan u = \frac{\sin u}{\cos u} = \frac{\frac{x}{a}}{\sqrt{1 - \frac{x^{2}}{a^{2}}}} = \frac{\frac{x}{a}}{\sqrt{\frac{a^{2} - x^{2}}{a^{2}}}} = \frac{\frac{x}{a}}{\frac{1}{a} \sqrt{a^{2} - x^{2}}} \\ \tan u = \frac{x}{a} \cdot \frac{a}{\sqrt{a^{2} - x^{2}}} = \frac{x}{\sqrt{a^{2} - x^{2}}} \\ \tan u = \frac{x}{\sqrt{a^{2} - x^{2}}} \end{aligned}

…(iv)
Now put the value of tan u and u from equation (iv) and (iii) in equation (ii), we get

I = \frac{x}{\sqrt{a^{2} - x^{2}}} - \sin^{- 1} (\frac{x}{a}) + c

Indefinite Integrals Excercise 18.13 Question 5

Answer:

\frac{1}{54} [\tan^{- 1} (\frac{x + 1}{3}) + \frac{3 (x + 1)}{x^{2} + 2 + 10}] + c

Hint: Use substitution method to solve this integral
Given:

\int \frac{1}{{(x^{2} + 2 x + 10)}^{2}} d x

Solution:

\begin{aligned} Let I = \int \frac{1}{{(x^{2} + 2 x + 10)}^{2}} d x \\ = \int \frac{1}{{(x^{2} + 2 x + 1 + 9)}^{2}} d x \\ = \int \frac{1}{{(x^{2} + 2 x \cdot 1 + 1) + 9}^{2}} d x \\ = \int \frac{1}{{(x + 1)^{2} + 9}^{2}} d x \end{aligned}

Putting

x + 1 = 3 \tan θ

…(i)

d x = 3 \sec^{2} θ d θ

Then

\begin{aligned} I = \int \frac{1}{{(3 \tan θ)^{2} + 9}^{2}} 3 \sec^{2} θ d θ \\ = \int \frac{3 \sec^{2} θ}{{9 (1 + \tan^{2} θ)}^{2}} d θ = \int \frac{3 \sec^{2} θ}{9^{2} {(\sec^{2} θ)}^{2}} d θ \end{aligned}

\begin{aligned} = \int \frac{3 \sec^{2} θ}{81 \sec^{4} θ} d θ [\begin{matrix} ∵ \sec^{2} θ - \tan^{2} θ = 1 \\ \Rightarrow 1 + \tan^{2} θ = \sec^{2} θ \end{matrix}] \\ = \int \frac{1}{27} \frac{1}{\sec^{2} θ} d θ = \int \frac{1}{27} \cos^{2} θ d θ [∵ \cos θ = \frac{1}{\sec θ}] \\ = \frac{1}{27} \int {\frac{1 + \cos 2 θ}{2}} d θ [\begin{matrix} ∵ 2 \cos^{2} A = 1 + \cos 2 A \\ \cos^{2} A = \frac{1 + \cos 2 A}{2} \end{matrix}] \end{aligned}

\begin{aligned} = \frac{1}{27} {\int \frac{1}{2} d θ + \int \frac{1}{2} \cos 2 θ d θ} \\ = \frac{1}{27} \times \frac{1}{2} \int θ^{0} d θ + \frac{1}{27} \times \frac{1}{2} \int \cos 2 θ d θ \\ = \frac{1}{54} \cdot \frac{θ^{0 + 1}}{0 + 1} + \frac{1}{54} \frac{\sin 2 θ}{2} + c [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c and \int \cos a x d x = \frac{\sin a x}{a} + c] \end{aligned}

= \frac{1}{54} θ + \frac{1}{108} \sin 2 θ + c

…(ii)
Also from (i)

\begin{aligned} x + 1 = 3 \tan θ \\ \tan θ = \frac{x + 1}{3} \Rightarrow θ = \tan^{- 1} (\frac{x + 1}{3}) \end{aligned}

… (iii)
And

\tan θ = \frac{x + 1}{3}

\begin{array}{ll} \Rightarrow \tan^{2} θ = {(\frac{x + 1}{3})}^{2} & [Squaring both sides] \\ \Rightarrow \frac{\sin^{2} θ}{\cos^{2} θ} = \frac{(x + 1)^{2}}{9} & [∵ \tan θ = \frac{\sin θ}{\cos θ}] \\ \Rightarrow \sin^{2} θ = \frac{(x + 1)^{2}}{9} \cdot \cos^{2} θ & [∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b] \end{array}

\begin{aligned} \Rightarrow \sin^{2} θ = \frac{(x^{2} + 2 x + 1) \cos^{2} θ}{9} \\ \Rightarrow \sin^{2} θ = \frac{(x^{2} + 2 x + 1)}{9} (1 - \sin^{2} θ) [\begin{matrix} \sin^{2} θ + \cos^{2} θ = 1 \\ \Rightarrow \cos^{2} θ = 1 - \sin^{2} θ \end{matrix}] \\ \Rightarrow \sin^{2} θ = \frac{(x^{2} + 2 x + 1)}{9} - (\frac{(x^{2} + 2 x + 1)}{9}) \sin^{2} θ \end{aligned}

\begin{aligned} \Rightarrow \frac{x^{2} + 2 x + 1}{9} = \sin^{2} θ + (\frac{x^{2} + 2 x + 1}{9}) \sin^{2} θ \\ \Rightarrow \frac{x^{2} + 2 x + 1}{9} = \sin^{2} θ {1 + (\frac{x^{2} + 2 x + 1}{9})} \\ \Rightarrow \frac{x^{2} + 2 x + 1}{9} = (\frac{9 + x^{2} + 2 x + 1}{9}) \sin^{2} θ = (\frac{x^{2} + 2 x + 10}{9}) \sin^{2} θ \end{aligned}

\begin{aligned} \Rightarrow \sin^{2} θ = \frac{x^{2} + 2 x + 1}{9} \times \frac{9}{x^{2} + 2 x + 10} = \frac{(x + 1)^{2}}{(x^{2} + 2 x + 10)} [∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b] \\ \Rightarrow \sin θ = \sqrt{\frac{(x + 1)^{2}}{x^{2} + 2 x + 10}} = \frac{(x + 1)}{\sqrt{x^{2} + 2 x + 10}} \dots (iv) \\ Since \sin^{2} θ + \cos^{2} θ = 1 \end{aligned}

So we can write

\begin{aligned} \Rightarrow \cos^{2} θ = 1 - \sin^{2} θ \\ \Rightarrow \cos θ = \sqrt{1 - \sin^{2} θ} \\ \begin{aligned} \Rightarrow \cos θ & = \sqrt{1 - {[\frac{(x + 1)}{\sqrt{x^{2} + 2 x + 10}}]}^{2}} \\ = \sqrt{\frac{x^{2} + 2 x + 10 - (x^{2} + 2 x + 1)}{x^{2} + 2 x + 10}} \end{aligned} [∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b] \end{aligned}

\begin{aligned} \Rightarrow \cos θ = \sqrt{\frac{x^{2} + 2 x + 10 - x^{2} - 2 x - 1}{x^{2} + 2 x + 10}} \\ \Rightarrow \cos θ = \sqrt{\frac{9}{x^{2} + 2 x + 10}} = \frac{3}{\sqrt{x^{2} + 2 x + 10}} . . . . (v) \end{aligned}

Also, we know

\sin 2 θ = 2 \sin θ \cos θ

\begin{aligned} \Rightarrow \sin 2 θ = 2 \frac{(x + 1)}{\sqrt{x^{2} + 2 x + 10}} \cdot \frac{3}{\sqrt{x^{2} + 2 x + 10}} [from (iv) and (v)] \\ \Rightarrow \sin 2 θ = \frac{2.3 (x + 1)}{x^{2} + 2 x + 10} = \frac{6 (x + 1)}{x^{2} + 2 x + 10} \cdot \cdot \cdot \cdot (v i) \end{aligned}

Now put the value of θ and sin2θ from equation (iii) and (vi) in (ii) we get

\begin{aligned} I = \frac{1}{54} θ + \frac{1}{108} \sin 2 θ + c \\ = \frac{1}{54} \tan^{- 1} (\frac{x + 1}{3}) + \frac{1}{108} \cdot \frac{6 (x + 1)}{(x^{2} + 2 x + 10)} + c \\ = \frac{1}{54} \tan^{- 1} (\frac{x + 1}{3}) + \frac{1}{54} \frac{3 (x + 1)}{(x^{2} + 2 x + 10)} + c \\ = \frac{1}{54} [\tan^{- 1} (\frac{x + 1}{3}) + \frac{3 (x + 1)}{(x^{2} + 2 x + 10)}] + c \end{aligned}

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