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

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

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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

Solution:

\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}} \\

 

I=\frac{\pi}{4}

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