If $n \geq 2$ is a natural number and $0 \lt \theta \lt \frac{\pi}{2}$, then $\int \frac{\left(\cos ^n…

If $n \geq 2$ is a natural number and $0 \lt \theta \lt \frac{\pi}{2}$, then $\int \frac{\left(\cos ^n \theta-\cos \theta\right)^{1 / n}}{\cos ^{n+1} \theta} \sin \theta d \theta=$
  1. $\frac{n}{n-1}\left(\cos ^{(1-n)} \theta-1\right)^2+c$
  2. $\frac{n}{(n+1)(1-n)}\left(\cos ^{(1-n)} \theta-1\right)^{1+\frac{1}{n}}+c$
  3. $\frac{n}{1-n}\left(\cos ^{(n-1)} \theta-1\right)^2+c$
  4. $\frac{n}{1-n^2}\left(1-\cos ^{(n-1)} \theta\right)^{(n+1) / n}$

Solution

$\mathrm{I}=\int \frac{\left(\cos ^n \theta-\cos \theta\right)^{\frac{1}{n}}}{\cos ^{n+1} \theta} \sin \theta d \theta$ $I=\int \frac{\cos \theta\left(1-\cos ^{1-n} \theta\right)^{\frac{1}{n}}}{\cos ^{n+1} \theta} \sin \theta d \theta$ $I=\int \frac{\left(1-\cos ^{1-n} \theta\right)^{\frac{1}{n}}}{\cos ^n \theta} \sin \theta d \theta$
Let $1-\cos ^{1-n} \theta=' t$ $\begin{aligned} & \Rightarrow(1-\mathrm{n}) \sin \theta \cos ^{-\mathrm{n}} \theta \mathrm{~d} \theta=\mathrm{dt} \\ & \therefore \mathrm{I}=\int \frac{t^{\frac{1}{n}}}{(1-n)} d t=\frac{\frac{1}{t^n+1}}{\left(\frac{1}{n}+1\right)(1-n)}=\frac{n t^{\frac{1+n}{n}}}{\left(1-n^2\right)} \\ & =\frac{n}{1-n^2}\left(1-\cos ^{(1-n)} \theta\right)^{\frac{n+1}{n}} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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