Integration by Parts Questions in English

(B) Let $I = \int \text{cosec}^{-1} \left( \sqrt{\frac{a + x}{x}} \right) dx$. Since $\text{cosec}^{-1} \sqrt{\frac{a+x}{x}} = \sin^{-1} \sqrt{\frac{x}{a+x}}$, let $x = a \tan^2 \theta$. Then $dx = 2a \tan \theta \sec^2 \theta d\theta$.
Substituting these, $I = \int \sin^{-1} \left( \sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}} \right) (2a \tan \theta \sec^2 \theta) d\theta = \int \sin^{-1} (\sin \theta) (2a \tan \theta \sec^2 \theta) d\theta = \int \theta (2a \tan \theta \sec^2 \theta) d\theta$.
Using integration by parts: $I = 2a \left[ \theta \int \tan \theta \sec^2 \theta d\theta - \int \left( 1 \cdot \int \tan \theta \sec^2 \theta d\theta \right) d\theta \right] = 2a \left[ \theta \cdot \frac{\tan^2 \theta}{2} - \int \frac{\tan^2 \theta}{2} d\theta \right] = a \theta \tan^2 \theta - a \int (\sec^2 \theta - 1) d\theta$.
$I = a \theta \tan^2 \theta - a (\tan \theta - \theta) + c = a \theta \tan^2 \theta - a \tan \theta + a \theta + c$.

(D) Let $I = \int 2x^{1/3} \sin x^{2/3} dx$.
Substitute $u = x^{2/3}$, then $du = \frac{2}{3} x^{-1/3} dx$, which implies $dx = \frac{3}{2} x^{1/3} du$.
Substituting into the integral: $I = \int 2x^{1/3} \sin(u) \cdot \frac{3}{2} x^{1/3} du = 3 \int x^{2/3} \sin(u) du = 3 \int u \sin u du$.
Using integration by parts $\int u \sin u du = u(-\cos u) - \int 1 \cdot (-\cos u) du = -u \cos u + \sin u$.
Thus, $I = 3[-u \cos u + \sin u] + c$.
Substituting $u = x^{2/3}$ back, $I = 3[-x^{2/3} \cos x^{2/3} + \sin x^{2/3}] + c$.

(A) Let $I = \int f(x)g(x) dx = \int \frac{\sin^{-1} x}{\sqrt{1-x^2}} e^{\sin^{-1} x} dx$.
Substitute $u = \sin^{-1} x$, then $du = \frac{1}{\sqrt{1-x^2}} dx$.
The integral becomes $I = \int u e^u du$.
Using integration by parts $\int u e^u du = u e^u - \int e^u du = u e^u - e^u + c$.
Substituting back $u = \sin^{-1} x$, we get $I = e^{\sin^{-1} x}(\sin^{-1} x - 1) + c$.

(A) Let $I = \int x^3 f(x^2) dx$.
Substitute $t = x^2$, then $dt = 2x dx$, which implies $x dx = \frac{1}{2} dt$.
Also, $x^2 = t$.
Substituting these into the integral: $I = \int x^2 f(x^2) (x dx) = \int t f(t) \frac{1}{2} dt = \frac{1}{2} \int t f(t) dt$.
Using integration by parts, $\int u dv = uv - \int v du$, let $u = t$ and $dv = f(t) dt$.
Then $du = dt$ and $v = \int f(t) dt = g(t)$.
Thus, $I = \frac{1}{2} [t g(t) - \int g(t) dt]$.
Substituting $t = x^2$ back, we get $I = \frac{1}{2} [x^2 g(x^2) - \int g(x^2) d(x^2)]$.

(C) Let $F(x) = \int f(x) dx$ and $G(x) = \int g(x) dx$. Then $F'(x) = f(x)$ and $G'(x) = g(x)$.
We want to find the expression for $F(x)G(x)$.
By the product rule for differentiation, $\frac{d}{dx}[F(x)G(x)] = F(x)G'(x) + F'(x)G(x) = F(x)g(x) + f(x)G(x)$.
Integrating both sides with respect to $x$, we get $F(x)G(x) = \int [f(x)G(x) + g(x)F(x)] dx$.
Substituting back $F(x) = \int f(x) dx$ and $G(x) = \int g(x) dx$, we get $F(x)G(x) = \int [f(x) \int g(x) dx + g(x) \int f(x) dx] dx$.