Test Index

NCERT Class XII Mathematics Chapter - - Solutions

Show Para  Hide Para 
Evaluate the definite integrals in Exercises 253 to 261.
© examsnet.com
Question : 259 of 271
Marks: +1, -0
∫0π/2sin⁡2xtan⁡−1(sin⁡x) dx\int\limits_{0}^{\pi/2} \sin 2x \tan^{-1}(\sin x) \, dx
Solution:  
Let I=∫0π/2sin⁡2xtan⁡−1(sin⁡x) dxI=\int\limits_{0}^{\pi/2} \sin 2x \tan^{-1}(\sin x) \, dx
⇒I=∫0π/22sin⁡xcos⁡xtan⁡−1(sin⁡x) dx\Rightarrow I=\int\limits_{0}^{\pi/2} 2\sin x \cos x \tan^{-1}(\sin x) \, dx
Put sin⁡x=t\sin x = t ⇒cos⁡x dx=dt\Rightarrow \cos x \, dx = dt
When x=0,t=0x = 0, t = 0 and when x=π/2,t=1x=\pi/2, t=1
∴I=2∫01ttan⁡−1t dt\therefore I=2\int\limits_{0}^{1} t \tan^{-1}t \, dt
=2[tan⁡−1t∫t dt−∫(ddt(tan⁡−1t)⋅∫t dt)dt]01=2\left[ \tan^{-1}t \int t \, dt - \int \left( \frac{d}{dt}(\tan^{-1}t) \cdot \int t \, dt \right) dt \right]_{0}^{1}
=2[tan⁡−1(t)t22−∫11+t2⋅t22 dt]01=2\left[ \tan^{-1}(t) \frac{t^{2}}{2} - \int \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} \, dt \right]_{0}^{1}
=2[t22tan⁡−1(t)−12∫1+t2−11+t2 dt]01=2\left[ \frac{t^{2}}{2}\tan^{-1}(t) - \frac{1}{2} \int \frac{1+t^{2}-1}{1+t^{2}} \, dt \right]_{0}^{1}
=[t2tan⁡−1(t)−∫(1−11+t2)dt]01= \left[ t^{2} \tan^{-1}(t) - \int \left(1 - \frac{1}{1+t^{2}}\right) dt \right]_{0}^{1}
=[t2tan⁡−1(t)−t+tan⁡−1t]01= \left[ t^{2} \tan^{-1}(t) - t + \tan^{-1}t \right]_{0}^{1}
=tan⁡−11−1+tan⁡−11= \tan^{-1}1 - 1 + \tan^{-1}1
=π4−1+π4=π2−1= \frac{\pi}{4} - 1 + \frac{\pi}{4} = \frac{\pi}{2} - 1
© examsnet.com
Go to Question:
Prev Question
Next Question
Advertisements