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  • Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 28 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 28 Maths Textbook Solution.

Answers (1)

Answer:  \frac{\pi}{4}-\frac{2}{3}

Given:  \int_{0}^{\frac{\pi}{4}} \tan ^{4} x d x

Hint: Use trigonometric identities and then integrate by parts

Solution:

\int_{0}^{\frac{\pi}{4}} \tan ^{4} x d x=\int_{0}^{\frac{\pi}{4}} \tan ^{2} x \tan ^{2} x d x

=\int_{0}^{\frac{\pi}{4}} \tan ^{2} x\left(\sec ^{2} x-1\right) d x \\

\begin{aligned} & &=\int_{0}^{\frac{\pi}{4}} \tan ^{2} x \sec ^{2} x d x-\int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} x-1\right) d x \end{aligned}

\begin{aligned} &p u t \\ & \end{aligned}

\tan x=t \\   (Differentiate w.r.t to x)

\sec ^{2} x d x=d t

\begin{aligned} &=\left(\frac{t^{3}}{3}\right)_{0}^{\frac{\pi}{4}}-\left(1-\frac{\pi}{4}\right)=\frac{1}{3}-1+\frac{\pi}{4} \\ & \end{aligned}

=\frac{\pi}{4}-\frac{2}{3}

 

Posted by

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