91Ó°ÊÓ

To evaluate \(\int x \sin x \cos ^{2} x d x\), we will use integration by parts. We can let u = x and dv = sinx*cos^2(x) dx: Step 1: Apply integration by parts \(\int u dv = uv - \int v du\) Step 2: Find du and v \(u = x \Rightarrow du = dx\) \(dv = \sin x \cos^2 x dx \Rightarrow v = \int \sin x \cos^2 x dx\) Step 3: Evaluate the integral of v To find v, we can perform a substitution \(t = \cos x \Rightarrow dt = -\sin x dx\) \(v = - \int t^2 dt = -\frac{1}{3}t^3 + C = -\frac{1}{3}\cos^3 x + C\) Step 4: Substitute back into the integration by parts formula \(\int x \sin x \cos^2 x dx = x\left(-\frac{1}{3}\cos^3 x\right) - \int -\frac{1}{3}\cos^3 x dx\) Step 5: Evaluate the remaining integral To evaluate the integral of \(\cos^3 x\), we can rewrite it as: \(\int \cos^3 x dx = \int (1-\sin^2 x)\cos x dx\) Using the substitution \(u = \sin x \Rightarrow du = \cos x dx\), we can perform the integration: \(\int (1-\sin^2 x)\cos x dx = \int (1-u^2) du\) This can be integrated easily: \(\int (1-u^2) du = u - \frac{1}{3}u^3 + C = \sin x - \frac{1}{3}\sin^3 x + C\) Step 6: Combine the results \(\int x \sin x \cos^2 x dx = -\frac{1}{3}x\cos^3 x - \sin x + \frac{1}{3}\sin^3 x + C\) For integral (ii):

To evaluate \(\int x \sec^2 x \tan x d x\), we will use the substitution method. Step 1: Apply substitution Let \(u=\tan x \Rightarrow du=\sec^2 x dx\) The given integral is then: \(\int x \sec^2 x \tan x d x = \int x u du\) Step 2: Apply integration by parts Let \(v=u\) and \(dw=xdx\) From integration by parts, we have: \(\int x u du = uv - \int v dw\) Find dv and w: \(dw = x dx \Rightarrow w = \frac{1}{2}x^2\) \(dv = u du \Rightarrow v = \frac{1}{2}u^2\) Now substitute back into the integration by parts formula: \(\int x u du = \frac{1}{2}x^2\left(\frac{1}{2}u^2\right) - \int \frac{1}{2}u^2 (\frac{1}{2}x^2 dx)\) Step 3: Reverse the original substitution Replace \(u\) by \(\tan x\): \(\int x \sec^2 x \tan x d x = \frac{1}{4}x^2 \tan^2 x - \frac{1}{4}\int \tan^2 x x^2 dx\) For integral (iii):

To evaluate \(\int x \cos x \cos 2 x d x\), we will use integration by parts along with the trigonometric identity \(\cos 2x = 1 - 2\sin^2 x\). Step 1: Apply integration by parts Let \(u=x\) and \(dv=\cos x\cos 2x dx\) Then the given integral is: \(\int u dv = uv - \int v du\) Step 2: Find du and v \(u = x \Rightarrow du = dx\) \(dv = \cos x \cos 2x dx \Rightarrow v = \int \cos x (1-2\sin^2 x) dx\) Step 3: Evaluate the integral of v To find \(v\): \(v = \int \cos x - 2\cos x\sin^2 x dx\) Now, we can evaluate each term separately: \(\int \cos x dx = \sin x + C\) \(\int -2\cos x\sin^2 x dx\), use the substitution method: \(t = \sin x \Rightarrow dt = \cos x dx\) \(v = -2 \int t^2 dt = -\frac{2}{3}t^3 + C = -\frac{2}{3}\sin^3 x + C\) Step 4: Substitute back into the integration by parts formula \(\int x\cos x\cos 2x dx = x(\sin x - \frac{2}{3}\sin^3 x) - \int (\sin x - \frac{2}{3}\sin^3 x) dx\) Step 5: Evaluate the remaining integral To evaluate the remaining integral: \(\int (\sin x - \frac{2}{3}\sin^3 x) dx\) Split the integral into two parts: \(\int \sin x dx - \frac{2}{3}\int \sin^3 x dx\) Now, we can evaluate each term separately: \(\int \sin x dx = -\cos x + C\) \(\int \sin^3 x dx\), use the substitution method: \(u = \cos x \Rightarrow du = -\sin x dx\) \(\int \sin^3 x dx = -\int (1-u^2)u du\) This can be integrated easily: \(-\int (1-u^2)u du = -u + \frac{1}{3}u^3 + C = -\cos x + \frac{1}{3}\cos^3 x + C\) Step 6: Combine the results \(\int x\cos x\cos 2x dx = x(-\cos x + \frac{1}{3}\cos^3 x) - (-\cos x + \frac{1}{3}\cos^3 x) dx\) This can be simplified to: \(\int x\cos x\cos 2x dx = -x\cos x + \frac{1}{3}x\cos^3 x + \cos x - \frac{1}{3}\cos^3 x + C\)