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  • Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 11 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 11 Maths Textbook Solution.

Answers (1)

Answer:

I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)+c

Hint:The given integral function can be converted into,

\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c

Given:

I=\int \frac{1}{\sin ^{4} x+\sin ^{2} x \cos ^{2} x+\cos ^{4} x}

Solution

\begin{aligned} I &=\int \frac{1}{\sin ^{4} x+\sin ^{2} x \cos ^{2} x+\cos ^{4} x} \\ &=\int \frac{\left(\tan ^{2 } x+1\right) \sec ^{2} x}{\tan ^{4} x+\tan ^{2} x+1} d x \end{aligned} (dividing N and D by \cos ^{4}x)

= \int \frac{t^{2}+1}{t^{4}+t^{2}+1}dt   where \tan x= t

=\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+1} d t         (Dividing N and D by t^{2})

putting\; t-\frac{1}{t}=z\; so\; that \left(1+\frac{1}{t^{2}}\right) d t=d \; z\; and \; t^{2}+\frac{1}{t^{2}}=z^{2}+2\\\\ \begin{aligned} I &=\int \frac{d z}{z^{2}+(\sqrt{3})^{2}} \\ &=\frac{1}{\sqrt{3}} \tan ^{-} \frac{z}{\sqrt{3}}+c \\ &=\frac{1}{\sqrt{3}} \tan ^{-}\left(\frac{t^{2}-1}{\sqrt{3} t}\right)+c \end{aligned}

The required value of the given integration is,

I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)+c

where c is an integrating constant.

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