The value of int frac{sqrt[n]{text{cosec}^{2} | vedclass

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(B) Let $I = \int \frac{\sqrt[n]{	ext{cosec}^{2}x^n - 1}}{x^{(1-n)}} dx$.Since $	ext{cosec}^{2}x^n - 1 = \cot^{2}x^n$, the integral becomes $I = \in

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$\frac{1}{n} \log|\sin x^n| + c$

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(B) Let $I = \int \frac{\sqrt[n]{	ext{cosec}^{2}x^n - 1}}{x^{(1-n)}} dx$.Since $	ext{cosec}^{2}x^n - 1 = \cot^{2}x^n$, the integral becomes $I = \int \frac{(\cot^{2}x^n)^{1/n}}{x^{1-n}} dx = \int \frac{(\cot x^n)^{2/n}}{x^{1-n}} dx$.Let $u = \cot x^n$. Then $du = -	ext{cosec}^{2}x^n \cdot n x^{n-1} dx$.This substitution does not simplify directly. Let us use $u = x^n$, then $du = n x^{n-1} dx$, so $dx = \frac{du}{n x^{n-1}} = \frac{du}{n u^{(n-1)/n}}$.$I = \int \frac{\sqrt[n]{\cot^2 u}}{x^{1-n}} \cdot \frac{du}{n x^{n-1}} = \int \frac{(\cot u)^{2/n}}{n} du$.Actually, the standard form is $\int \frac{\cot x^n}{x^{1-n}} dx$. Let $u = x^n$, $du = n x^{n-1} dx$. Then $dx = \frac{du}{n x^{n-1}}$.$I = \int \frac{\cot u}{x^{1-n}} \cdot \frac{du}{n x^{n-1}} = \frac{1}{n} \int \cot u du = \frac{1}{n} \log|\sin u| + c = \frac{1}{n} \log|\sin x^n| + c$.

The value of $\int \frac{\sqrt[n]{\text{cosec}^{2}x^n - 1}}{x^{(1-n)}} dx$ is...

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