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  • Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 15

Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 15

Answers (1)

Answer:

        \frac{1}{6} \log \left|\frac{1+\tan ^{3} x}{1-\tan ^{3} x}\right|+C

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

        \int \frac{\tan ^{2} x \sec ^{2} x}{1-\tan ^{6} x} d x

Solution:

Let\: \: I=\int \frac{\tan ^{2} x \sec ^{2} x}{1-\tan ^{6} x} d x

             =\int \frac{\tan ^{2} x \sec ^{2} x}{1-(\tan ^{3} x)^{2}} d x

\begin{aligned} &\text { Put } \tan ^{3} x=t \Rightarrow 3 \tan ^{2} x \cdot \sec ^{2} x d x=d t \\ &\text { Then } I=\int \frac{1}{1-t^{2}} \cdot \frac{d t}{3} \\ &=\frac{1}{3} \int \frac{1}{1^{2}-t^{2}} d t \\ &=\frac{1}{3} \cdot \frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|+C \quad\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right] \\ &=\frac{1}{6} \log \left|\frac{1+\tan ^{3} x}{1-\tan ^{3} x}\right|+C \quad\left[\because t=\tan ^{3} x\right] \end{aligned}

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Gurleen Kaur

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