91Ó°ÊÓ

Let's simplify the given integral using the trigonometric identity \(\cosec{x} = \frac{1}{\sin{x}}\). Then the integral becomes: \(\int \cos ^{5} x \operatorname{cosec}^{2} x d x = \int \cos ^{5} x \frac{1}{\sin ^{2} x} d x\) Now, let's perform substitution. Let \(u = \sin{x}\), so \(du = \cos{x} dx\). The integral becomes: \(\int \frac{\cos ^{4} x}{u^{2}} du\) Using the identity \(1 - \sin ^{2}(x) = \cos ^{2}(x)\), we can rewrite the integral as: \(\int \frac{(1 - u^{2})^{2}}{u^{2}} du\) Now we can expand and integrate each term separately: \(\int \frac{1 - 2u^{2} + u^{4}}{u^{2}} du = \int \frac{1}{u^{2}} du - 2\int u^{(0)} du + \int u^{2} du\) We can then integrate each term: \(\int \frac{1}{u^{2}} du - 2\int u^{(0)} du + \int u^{2} du = -\frac{1}{u} - 2u + \frac{u^{3}}{3} + C\) Now substitute back \(u = \sin{x}\): \(-\frac{1}{\sin{x}} - 2\sin{x} + \frac{\sin^{3}{x}}{3} + C\) (i) Integral: \(\int \cos ^{5} x \operatorname{cosec}^{2} x d x = -\frac{1}{\sin{x}} - 2\sin{x} + \frac{\sin^{3}{x}}{3} + C\)

To evaluate this integral, we will make use of substitution again. Let \(u = \sin{x}\), so \(du = \cos{x} dx\). The integral becomes: \(\int \sqrt{\frac{1 - u^2}{u^5}} du\) Now we can write it as: \(\int \frac{\sqrt{1 - u^2}}{u^{\frac{5}{2}}} du\) To solve this integral, we can transform it to the following trigonometric form, which is easier to evaluate. Let \(u = \cos{\theta}\), so \(du = -\sin{\theta} d\theta\) and \(\sqrt{1 - u^2} = \sin{\theta}\). The integral is then: \(-\int \frac{\sin ^2{\theta}}{\cos ^{\frac{5}{2}}{\theta}} d\theta\) Now, let's use integration by parts, by setting \(I(u) = \sin{\theta}\) and \(I'(dv) = \cot ^{\frac{1}{2}}{\theta}\sin{\theta} d\theta\). Now, differentiate \(I(u)\) to find \(I'(du)\): \(I'(du) = \cos{\theta} d\theta\) Integrate \(I'(dv)\) to find \(I(v)\): \(I(v) = \int \cot ^{\frac{1}{2}}{\theta}\sin{\theta} d\theta = \frac{2}{3}\cot ^{\frac{3}{2}}{\theta} + C\) Then, by integration by parts: \(uv - \int vI'(du) = \sin{\theta}\cdot\frac{2}{3}\cot ^{\frac{3}{2}}{\theta} - \int \frac{2}{3}\cot ^{\frac{1}{2}}{\theta}\cos{\theta} d\theta\) The integral on the right is just \(\frac{2}{3}I\), so we can rearrange to find the solution as: \(I = -\frac{3}{5}\sin{\theta}\cot ^{\frac{3}{2}}{\theta} + C\) Now substitute back \(\theta = \arccos{u}\), and then substitute back \(u = \sin{x}\): (ii) Integral: \(\int \sqrt{\frac{\cos x}{\sin ^{5} x}} d x = -\frac{3}{5}\sin{x}\csc ^{\frac{3}{2}}{x} \cot ^{\frac{3}{2}}{x} + C\)

To begin with, change \(\sec^2{x}\) into \(\frac{1}{\cos^2{x}}\) and make use of identity \(\sin2x = 2 \sin x \cos x\). \(\int \frac{\sec ^{2} x}{\sin 2 x} d x = \int \frac{1}{\cos^2{x} (2 \sin x \cos x)} d x\) Now, the fraction simplifies as: \(\int \frac{1}{2 \sin x \cos^3{x}} d x\) Now let \(u = \cos x\), so \(du = -\sin x d x\). The integral becomes: \(-\frac{1}{2}\int \frac{1}{u^3} du\) Integrating, we get: \(-\frac{1}{2} \cdot -\frac{1}{2} u^{-2} + C = \frac{1}{4} u^{-2} + C\) Now substitute back \(u = \cos{x}\): (iii) Integral: \(\int \frac{\sec ^{2} x}{\sin 2 x} d x = \frac{1}{4} \sec^2{x} + C\)

To evaluate this integral, perform substitution. Let \(u = \frac{x}{2}\), so \(du = \frac{1}{2} dx\). The integral becomes: \(\int \frac{2 du}{\sin u \sqrt{\cos^3 u}}\) Now perform another substitution, and let \(v = \cos u\), so \(dv = -\sin u du\). The integral becomes: \(-2\int \frac{dv}{\sqrt{v^3}}\) Integrate the function: \(-2\int v^{-\frac{3}{2}} dv = -2 \left( -\frac{2}{\sqrt{1-v^2}}\right) + C = \frac{4}{\sqrt{1 - v^2}} + C\) Now substitute back \(v = \cos u\) and then substitute back \(u = \frac{x}{2}\): (iv) Integral: \(\int \frac{d x}{\sin \frac{x}{2} \sqrt{\cos ^{3} \frac{x}{2}}} = \frac{4}{\sqrt{1 - \cos^2{\frac{x}{2}}}} + C = \frac{4}{\sin{\frac{x}{2}}} + C\)