Exercise 7.6 Practice

Step 1: Apply Integration by Parts formula: $\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$.

Step 2: Choose $u$ and $v$ using ILATE rule. Here $x$ is Algebraic and $\sin x$ is Trigonometric. So, $u=x$ and $v=\sin x$.

Step 3: Substitute into formula:
$= x \int \sin x \, dx - \int \left[\frac{d}{dx}(x) \int \sin x \, dx\right] \, dx$
$= x(-\cos x) - \int 1 \cdot (-\cos x) \, dx$

Step 4: Integrate the remaining part:
$= -x \cos x + \int \cos x \, dx$
$= -x \cos x + \sin x + C$

Step 1: Let $u=x$ and $v=\sin 3x$ (ILATE).

Step 2: Apply formula:
$= x \int \sin 3x \, dx - \int \left[\frac{d}{dx}(x) \int \sin 3x \, dx\right] \, dx$
$= x\left(-\frac{\cos 3x}{3}\right) - \int 1 \cdot \left(-\frac{\cos 3x}{3}\right) \, dx$

Step 3: Simplify and integrate:
$= -\frac{x}{3}\cos 3x + \frac{1}{3} \int \cos 3x \, dx$
$= -\frac{x}{3}\cos 3x + \frac{1}{3} \left(\frac{\sin 3x}{3}\right) + C$

Step 1: Let $u=x^2$ and $v=e^x$.

Step 2: Apply formula:
$= x^2 e^x - \int 2x e^x \, dx$

Step 3: Apply parts again for $\int x e^x \, dx$ (Let $u=x, v=e^x$):
$\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x$

Step 4: Combine results:
$= x^2 e^x - 2(x e^x - e^x) + C$
$= e^x(x^2 - 2x + 2) + C$

Step 1: According to ILATE, Logarithmic comes before Algebraic. So, $u=\log x$ and $v=x$.

Step 2: Apply formula:
$= \log x \int x \, dx - \int \left[\frac{d}{dx}(\log x) \int x \, dx\right] \, dx$
$= \log x \left(\frac{x^2}{2}\right) - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx$

Step 3: Simplify:
$= \frac{x^2}{2}\log x - \frac{1}{2} \int x \, dx$
$= \frac{x^2}{2}\log x - \frac{1}{2} \left(\frac{x^2}{2}\right) + C$

Step 1: Let $u=\log 2x$ and $v=x$.

Step 2: Apply formula:
$= \log 2x \left(\frac{x^2}{2}\right) - \int \frac{1}{2x} \cdot 2 \cdot \frac{x^2}{2} \, dx$
$= \frac{x^2}{2}\log 2x - \int \frac{x}{2} \, dx$

Step 3: Integrate:
$= \frac{x^2}{2}\log 2x - \frac{1}{2} \left(\frac{x^2}{2}\right) + C$

Step 1: Let $u=\log x$ and $v=x^2$.

Step 2: Apply formula:
$= \log x \left(\frac{x^3}{3}\right) - \int \frac{1}{x} \cdot \frac{x^3}{3} \, dx$
$= \frac{x^3}{3}\log x - \frac{1}{3} \int x^2 \, dx$

Step 3: Integrate:
$= \frac{x^3}{3}\log x - \frac{1}{3} \left(\frac{x^3}{3}\right) + C$

Step 1: Let $u=\sin^{-1} x$ and $v=x$.

Step 2: Apply formula:
$= \sin^{-1} x \left(\frac{x^2}{2}\right) - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx$
$= \frac{x^2}{2}\sin^{-1} x - \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx$

Step 3: Solve integral $I_1 = \int \frac{x^2}{\sqrt{1-x^2}} \, dx$.
Rewrite $x^2$ as $1 - (1-x^2) - 1$:
$I_1 = \int \frac{1-(1-x^2)-1}{\sqrt{1-x^2}} \, dx = \int \frac{-(1-x^2)+1}{\sqrt{1-x^2}} \, dx$
$= -\int \sqrt{1-x^2} \, dx + \int \frac{1}{\sqrt{1-x^2}} \, dx$
$= -\left[\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x\right] + \sin^{-1} x$

Step 4: Substitute back:
$= \frac{x^2}{2}\sin^{-1} x - \frac{1}{2}\left[-\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x\right] + C$
$= \frac{x^2}{2}\sin^{-1} x + \frac{x}{4}\sqrt{1-x^2} - \frac{1}{4}\sin^{-1} x + C$

Step 1: Let $u=\tan^{-1} x$ and $v=x$.

Step 2: Apply formula:
$= \frac{x^2}{2}\tan^{-1} x - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx$

Step 3: Simplify integral:
$\int \frac{x^2}{1+x^2} \, dx = \int \frac{1+x^2-1}{1+x^2} \, dx = \int \left(1 - \frac{1}{1+x^2}\right) \, dx$
$= x - \tan^{-1} x$

Step 4: Combine:
$= \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}(x - \tan^{-1} x) + C$

Step 1: Let $u=\cos^{-1} x$ and $v=x$.

Step 2: Apply formula:
$= \frac{x^2}{2}\cos^{-1} x - \int \frac{-1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx$
$= \frac{x^2}{2}\cos^{-1} x + \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx$

Step 3: Using result from Q7 for $\int \frac{x^2}{\sqrt{1-x^2}} \, dx$:
$= \frac{x^2}{2}\cos^{-1} x + \frac{1}{2} \left(-\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x\right) + C$
Note: $\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$. We can group constants.

Step 1: Let $\sin^{-1} x = \theta \Rightarrow x = \sin \theta \Rightarrow dx = \cos \theta \, d\theta$.
Integral becomes $\int \theta^2 \cos \theta \, d\theta$.

Step 2: Parts with $u=\theta^2, v=\cos \theta$:
$= \theta^2 \sin \theta - \int 2\theta \sin \theta \, d\theta$

Step 3: Parts again for $\int \theta \sin \theta \, d\theta$:
$= \theta(-\cos \theta) - \int 1(-\cos \theta) \, d\theta = -\theta \cos \theta + \sin \theta$

Step 4: Combine:
$= \theta^2 \sin \theta - 2(-\theta \cos \theta + \sin \theta) + C$
$= x(\sin^{-1} x)^2 + 2\sqrt{1-x^2}\sin^{-1} x - 2x + C$

Step 1: Let $\cos^{-1} x = t \Rightarrow -\frac{1}{\sqrt{1-x^2}} \, dx = dt$.
Also $x = \cos t$.

Step 2: Integral becomes $\int t \cos t (-dt) = -\int t \cos t \, dt$.

Step 3: Integrate by parts:
$= -(t \sin t - \int \sin t \, dt) = -(t \sin t + \cos t) + C$

Step 4: Substitute back ($t=\cos^{-1}x, \sin t = \sqrt{1-x^2}$):
$= -(\sqrt{1-x^2}\cos^{-1} x + x) + C$

Step 1: Let $u=x$ and $v=\sec^2 x$.

Step 2: Apply formula:
$= x \tan x - \int 1 \cdot \tan x \, dx$

Step 3: Integrate $\tan x$:
$= x \tan x - \log|\sec x| + C$

Step 1: Write as $1 \cdot \tan^{-1} x$. Let $u=\tan^{-1} x, v=1$.

Step 2: Apply formula:
$= x \tan^{-1} x - \int \frac{1}{1+x^2} \cdot x \, dx$

Step 3: Solve integral $\int \frac{x}{1+x^2} \, dx$:
Let $1+x^2=t \Rightarrow 2x dx = dt$. Integral is $\frac{1}{2}\log|t|$.

Step 1: Let $u=(\log x)^2$ and $v=x$.

Step 2: Apply formula:
$= (\log x)^2 \frac{x^2}{2} - \int 2\log x \cdot \frac{1}{x} \cdot \frac{x^2}{2} \, dx$
$= \frac{x^2}{2}(\log x)^2 - \int x \log x \, dx$

Step 3: Use result from Q4 for $\int x \log x \, dx$:
$= \frac{x^2}{2}(\log x)^2 - \left(\frac{x^2}{2}\log x - \frac{x^2}{4}\right) + C$

Step 1: Let $u=\log x$ and $v=x^2+1$.

Step 2: Apply formula:
$= \log x \left(\frac{x^3}{3} + x\right) - \int \frac{1}{x} \left(\frac{x^3}{3} + x\right) \, dx$
$= \left(\frac{x^3}{3} + x\right)\log x - \int \left(\frac{x^2}{3} + 1\right) \, dx$

Step 3: Integrate:
$= \left(\frac{x^3}{3} + x\right)\log x - \left(\frac{x^3}{9} + x\right) + C$

Step 1: Use the form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$.

Step 2: Let $f(x) = \sin x$, then $f'(x) = \cos x$.

Step 3: The integral matches the form.

Step 1: Rewrite numerator $x = (x+1) - 1$.
Integral becomes $\int e^x \left[\frac{x+1}{(1+x)^2} - \frac{1}{(1+x)^2}\right] \, dx$
$= \int e^x \left[\frac{1}{1+x} + \frac{-1}{(1+x)^2}\right] \, dx$

Step 2: Let $f(x) = \frac{1}{1+x}$, then $f'(x) = -\frac{1}{(1+x)^2}$.

Step 3: Using standard form:

Step 1: Simplify the trigonometric term:
$\frac{1+\sin x}{1+\cos x} = \frac{1+2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{1}{2}\sec^2(x/2) + \tan(x/2)$

Step 2: Integral is $\int e^x [\tan(x/2) + \frac{1}{2}\sec^2(x/2)] \, dx$.

Step 3: Let $f(x) = \tan(x/2)$, then $f'(x) = \frac{1}{2}\sec^2(x/2)$.

Step 1: Identify $f(x) = \frac{1}{x}$.

Step 2: Then $f'(x) = -\frac{1}{x^2}$.

Step 3: Matches form $\int e^x [f(x) + f'(x)] \, dx$.

Step 1: Rewrite $x-3 = (x-1) - 2$.
$\frac{x-3}{(x-1)^3} = \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3}$

Step 2: Let $f(x) = \frac{1}{(x-1)^2} = (x-1)^{-2}$.
Then $f'(x) = -2(x-1)^{-3} = -\frac{2}{(x-1)^3}$.

Step 3: Matches standard form.

Step 1: Let $I = \int e^{2x} \sin x \, dx$. Use parts with $u=\sin x, v=e^{2x}$.
$I = \sin x \frac{e^{2x}}{2} - \int \cos x \frac{e^{2x}}{2} \, dx$

Step 2: Apply parts again on $\int e^{2x} \cos x \, dx$:
$\int e^{2x} \cos x \, dx = \cos x \frac{e^{2x}}{2} - \int (-\sin x) \frac{e^{2x}}{2} \, dx$

Step 3: Substitute back:
$I = \frac{e^{2x}}{2}\sin x - \frac{1}{2}\left[\frac{e^{2x}}{2}\cos x + \frac{1}{2}I\right]$
$I = \frac{e^{2x}}{2}\sin x - \frac{e^{2x}}{4}\cos x - \frac{1}{4}I$

Step 4: Solve for $I$:
$\frac{5}{4}I = \frac{e^{2x}}{4}(2\sin x - \cos x)$

Step 1: Put $x = \tan \theta \Rightarrow dx = \sec^2 \theta \, d\theta$.
$\sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1} x$.

Step 2: Integral becomes $\int 2\tan^{-1} x \, dx$.

Step 3: Using result from Q13:
$= 2\left(x \tan^{-1} x - \frac{1}{2}\log(1+x^2)\right) + C$

Step 1: Let $x^3 = t \Rightarrow 3x^2 \, dx = dt \Rightarrow x^2 \, dx = \frac{dt}{3}$.

Step 2: Integral becomes $\frac{1}{3} \int e^t \, dt = \frac{1}{3}e^t + C$.

Step 1: Expand: $e^x (\sec x + \sec x \tan x)$.

Step 2: Let $f(x) = \sec x$, then $f'(x) = \sec x \tan x$.

Step 3: Matches standard form.