Question

Evaluate the following integrals: (i) \(\int \cos ^{5} x \operatorname{cosec}^{2} x d x\) (ii) \(\int \sqrt{\frac{\cos x}{\sin ^{5} x}} d x\) (iii) \(\int \frac{\sec ^{2} x}{\sin 2 x} d x\) (iv) \(\int \frac{d x}{\sin \frac{x}{2} \sqrt{\cos ^{3} \frac{x}{2}}}\)

Short answer

In summary, the evaluated integrals are as follows: (i) \(\int \cos ^{5} x \operatorname{cosec}^{2} x d x = -\operatorname{cosec}(x) + 2\sin(x) + \frac{\sin^3(x)}{3} + C\) (ii) \(\int \sqrt{\frac{\cos x}{\sin ^{5} x}} d x = -(\sin x)^{-\frac{1}{2}}(1-\sin^2 x)^{\frac{1}{2}} + C\) (iii) \(\int \frac{\sec ^{2} x}{\sin 2 x} d x = \frac{1}{4} \left(\ln|\tan(x)| + \frac{\tan^2(x)}{2}\right) + C\) (iv) \(\int \frac{dx}{\sin \frac{x}{2}\sqrt{\cos^{3} \frac{x}{2}}} = -\ln\left|\frac{\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}}\right| + C\)

Step-by-step solution

(i) Evaluate \(\int \cos ^{5} x \operatorname{cosec}^{2} x d x\)

To solve this integral, first rewrite the integrand using a trigonometric identity. Remember that \(\operatorname{cosec}(x) = \frac{1}{\sin(x)}\), so we have: \(\int \cos ^{5} x \left(\frac{1}{\sin^{2}x}\right) dx\). Now, let's use substitution: let \(t = \sin(x)\), which means \(dt = \cos(x) dx\). Also, rewrite \(\cos^{4}(x)\) using the Pythagorean identity: \(\cos^{2}(x) = 1 - \sin^2(x) = 1 - t^2\). So, the integral becomes: \[\int \cos ^{5} x \left(\frac{1}{\sin^{2}x}\right) dx = \int (1-t^2)^2 \left(\frac{1}{t^2}\right) dt\] Now, just expand the terms, and integrate term by term. \[\int (1-t^2)^2 \left(\frac{1}{t^2}\right) dt = \int \left(\frac{1}{t^2} - 2 + t^2\right) dt\] \[\int \left(\frac{1}{t^2} - 2 + t^2\right) dt = \int \frac{1}{t^2} dt - 2\int 1 dt + \int t^2 dt\] Now, evaluate each integral: \- \(\int \frac{1}{t^2} dt = -\frac{1}{t}\) \- \(2\int 1 dt = 2t\) \- \(\int t^2 dt = \frac{t^3}{3}\) So, our final solution is: \[-\frac{1}{t} + 2t + \frac{t^3}{3} + C\] Now, substitute back \(t = \sin(x)\): \[-\frac{1}{\sin(x)} + 2\sin(x) + \frac{\sin^3(x)}{3} + C = -\operatorname{cosec}(x) + 2\sin(x) + \frac{\sin^3(x)}{3} + C\] So, \(\int \cos ^{5} x \operatorname{cosec}^{2} x d x = -\operatorname{cosec}(x) + 2\sin(x) + \frac{\sin^3(x)}{3} + C\)

(ii) Evaluate \(\int \sqrt{\frac{\cos x}{\sin ^{5} x}} d x\)

To solve this integral, we can apply a substitution. We can rewrite the integral as: \(\int \left(\frac{\cos x}{\sin ^{5} x}\right)^{1/2} d x\). Now, let \(u = \sin x\), which means \(du = \cos x dx\). The integral then becomes: \[ \int \left(\frac{1-u^2}{u^5}\right)^{1/2} du = \int \frac{\sqrt{1-u^2}}{u^{\frac{5}{2}}} du\] Now, we can apply another substitution: \(v = u^2 \implies dv = 2u du\) \[ \frac{1}{2}\int \frac{\sqrt{1-v}}{v^{\frac{3}{2}}} dv\] Integrating the remaining integral: \[ \frac{1}{2}\int \frac{\sqrt{1-v}}{v^{\frac{3}{2}}} dv = \frac{1}{2}\left[-2(1-v)^{\frac{1}{2}}v^{-\frac{1}{2}} + C\right]\] Now, substitute back \(v = u^2\) and \(u = \sin x\): \[ -(\sin x)^{-\frac{1}{2}}(1-\sin^2 x)^{\frac{1}{2}} + C\] So, \(\int \sqrt{\frac{\cos x}{\sin ^{5} x}} d x = -(\sin x)^{-\frac{1}{2}}(1-\sin^2 x)^{\frac{1}{2}} + C\).

(iii) Evaluate \(\int \frac{\sec ^{2} x}{\sin 2 x} d x\)

To solve this integral, first use the trigonometric identity \(\sec^2(x) = 1 + \tan^2(x)\). Also, note that \(\sin 2x = 2 \sin x \cos x\). The integral becomes: \[ \int \frac{1 + \tan^2 x}{2 \sin x \cos x} dx = \frac{1}{2} \int \frac{1 + \tan^2 x}{\sin x \cos x} dx\] Now, use substitution: let \(w = \tan(x)\), which means \(dw = \sec^2(x) dx\). Also, use the identity \(1 + \tan^2 x = \sec^2 x\): \[ \frac{1}{2} \int \frac{1 + w^2}{2w(1 + w^2)^{1/2}} dw = \frac{1}{4}\int \frac{1 + w^2}{w} dw\] Now, split the fraction: \[\frac{1}{4}\int \left(\frac{1}{w} + w\right) dw = \frac{1}{4} \left(\int \frac{1}{w} dw + \int w dw\right)\] Evaluate each integral: \- \(\int \frac{1}{w} dw = \ln|w|\) \- \(\int w dw = \frac{w^2}{2}\) So, our final solution is: \[\frac{1}{4} \left(\ln|w| + \frac{w^2}{2}\right) + C\] Now, substitute back \(w = \tan(x)\): \[\frac{1}{4} \left(\ln|\tan(x)| + \frac{\tan^2(x)}{2}\right) + C\] So, \(\int \frac{\sec ^{2} x}{\sin 2 x} d x = \frac{1}{4} \left(\ln|\tan(x)| + \frac{\tan^2(x)}{2}\right) + C\).

(iv) Evaluate \(\int \frac{dx}{\sin \frac{x}{2}\sqrt{\cos^{3} \frac{x}{2}}}\)

For this integral, use substitution: let \(y = \cos\frac{x}{2}\), which means \(-2 \sin\frac{x}{2} dy = dx\). The integral then becomes: \[ -2\int \frac{1}{\sin\frac{x}{2}\sqrt{y^3}} dy = -2 \int \frac{dy}{y\sqrt{(1 - y^2)y}}\] Now, use another substitution: \(z = y^2 \implies dz = 2y dy\) \[ -\int \frac{dz}{z\sqrt{1 - z}\sqrt{z}} = -\int \frac{dz}{z\sqrt{z(1-z)}}\] Now, split the fraction: \< -\int \left(\frac{1}{z} - \frac{1}{1-z} \right)dz = -(\int \frac{1}{z}dz - \int \frac{1}{1-z}dz)> Evaluate each integral: \- \(\int \frac{1}{z} dz = \ln|z|\) \- \(-\int \frac{1}{1-z} dz = \ln|1-z|\) So, our final solution is: \[-\ln\left|\frac{z}{1-z}\right| + C\] Now, substitute back \(z = y^2\) and \(y = \cos\frac{x}{2}\): \[-\ln\left|\frac{\cos^2\frac{x}{2}}{1-\cos^2\frac{x}{2}}\right| + C = -\ln\left|\frac{\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}}\right| + C\] So, \(\int \frac{dx}{\sin \frac{x}{2} \sqrt{\cos^{3} \frac{x}{2}}} = -\ln\left|\frac{\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}}\right| + C\).