JEE Main 2019 — Mathematics Calculus

$$\int \frac{{({\mathrm{sin}}^{n}\theta -\mathrm{sin}\theta )}^{\frac{1}{n}}\mathrm{cos}\theta }{{\mathrm{sin}}^{n+1}\theta }d\theta$$∫sinn+1θ(sinnθ−sinθ)n1​cosθ​dθ

Put, $\mathrm{sin}\theta =t\Rightarrow \mathrm{cos}\theta d\theta =dt$sinθ=t⇒cosθdθ=dt

$$=\begin{matrix}\int \frac{{({t}^{n}-t)}^{\frac{1}{n}}dt}{{t}^{n+1}} & \end{matrix}$$=∫tn+1(tn−t)n1​dt​​​

$$=\int \frac{t{(1-\frac{1}{{t}^{n-1}})}^{\frac{1}{n}}}{{t}^{n+1}}dt$$=∫tn+1t(1−tn−11​)n1​​dt

$$= \int \frac{{(1-\frac{1}{{t}^{n-1}})}^{\frac{1}{n}}}{{t}^{n}}dt$$=∫tn(1−tn−11​)n1​​dt

Put $1-\frac{1}{{t}^{n-1}}=z$1−tn−11​=z

$$\Rightarrow \frac{(n-1)}{{t}^{n}}dt=dz$$⇒tn(n−1)​dt=dz

$$\Rightarrow I=\frac{1}{n-1}\int {z}^{\frac{1}{n}}dz$$⇒I=n−11​∫zn1​dz

Using $\displaystyle \int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+c$∫xndx=n+1xn+1​+c

$$\Rightarrow I=\frac{{z}^{\frac{1}{n}+1}}{(\frac{1}{n}+1)(n-1)}+c$$⇒I=(n1​+1)(n−1)zn1​+1​+c

$\Rightarrow I=\frac{n{(1-{t}^{1-n})}^{\frac{1}{n}+1}}{{n}^{2}-1}+c,$⇒I=n2−1n(1−t1−n)n1​+1​+c, where $c$c is the constant of integration.

$$\Rightarrow I=\frac{n}{{n}^{2}-1}{(1-\frac{1}{{\mathrm{sin}}^{n-1}\theta })}^{\frac{n+1}{n}}+c.$$⇒I=n2−1n​(1−sinn−1θ1​)nn+1​+c.