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Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 40

Answers (1)

Answer:  \frac{\pi}{6}

Hint: To solve this equation we will use  \int f\left ( x \right )dx  formula

Given:   \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{3} x} d x


\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{3}\left(\frac{\pi}{2}-x\right)} d x \\ & \end{aligned}                                        \left[\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]

\because \tan \left(\frac{\pi}{2}-x\right)=\cot x=\frac{1}{\tan x}

I=\int_{0}^{\frac{\pi}{2}} \frac{\tan ^{3} x}{\tan ^{3} x+1} d x    …………. (1)

I=\int_{0}^{\frac{\pi}{2}} \frac{1}{\tan ^{3} x+1} d x     ……………. (2)

2 I=\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\tan ^{3} x}{\tan ^{3} x+1}\right) d x


2 I=[x]_{0}^{\frac{\pi}{2}} \\



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