Question bank/JEE Main/Mathematics

JEE MainMathematicsToughIntegration

Let n ≥ 2 be a natural number and
0<θ<π/2.0 < \theta < \pi/2. Then ∫(sin⁡nθ−sinθ)1ncosθsin⁡n+1θdθ\int\frac{\left( \sin^{n}\theta - sin\theta \right)^{\frac{1}{n}}cos\theta}{\sin^{n + 1}\theta}d\theta is
equal to

Show answer and solution
∫(sin⁡nθ−sinθ)1ncosθsin⁡n+1θdθ\int\frac{\left( \sin^{n}\theta - sin\theta \right)^{\frac{1}{n}}cos\theta}{\sin^{n + 1}\theta}d\theta
=∫(tn−t)1ndttn+1= \int\frac{\left( t^{n} - t \right)^{\frac{1}{n}}dt}{t^{n + 1}} (Put
sinθ=tsin\theta = t)
=∫t(1−1tn−1)1ntn+1dt=∫(1−1tn−1)1ntndt= \int\frac{t\left( 1 - \frac{1}{t^{n - 1}} \right)^{\frac{1}{n}}}{t^{n + 1}}dt = \int\frac{\left( 1 - \frac{1}{t^{n - 1}} \right)^{\frac{1}{n}}}{t^{n}}dt
Put 1−1tn−1=z⇒(n−1)tn1 - \frac{1}{t^{n - 1}} = z \Rightarrow \frac{(n - 1)}{t^{n}} dt =
dz
⇒I=1n−1∫z1ndz=z1n+1(1n+1)(n−1)+c=n(1−t1−n)1n+1n2−1+c\Rightarrow I = \frac{1}{n - 1}\int z^{\frac{1}{n}}dz = \frac{z^{\frac{1}{n} + 1}}{\left( \frac{1}{n} + 1 \right)(n - 1)} + c = \frac{n\left( 1 - t^{1 - n} \right)^{\frac{1}{n} + 1}}{n^{2} - 1} + c
=nn2−1(1−1sin⁡n−1θ)n+1n+c= \frac{n}{n^{2} - 1}\left( 1 - \frac{1}{\sin^{n - 1}\theta} \right)^{\frac{n + 1}{n}} + c

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans