Integration by substitution Questions in English

(A) Let $I = \int \frac{\sin^3 x}{(\cos^4 x + 3 \cos^2 x + 1) \tan^{-1}(\sec x + \cos x)} dx$.
Substitute $u = \sec x + \cos x$. Then $du = (\sec x \tan x - \sin x) dx = (\frac{\sin x}{\cos^2 x} - \sin x) dx = \sin x (\frac{1 - \cos^2 x}{\cos^2 x}) dx = \sin x \tan^2 x dx$.
Note that $\sin^3 x dx = \sin x (1 - \cos^2 x) dx$. This does not simplify directly. Let us rewrite the denominator term: $\cos^4 x + 3 \cos^2 x + 1 = \cos^2 x (\cos^2 x + 3 + \sec^2 x) = \cos^2 x ((\sec x + \cos x)^2 + 1)$.
Thus, $I = \int \frac{\sin x (1 - \cos^2 x)}{\cos^2 x (u^2 + 1) \tan^{-1} u} dx = \int \frac{\sin x (\sec^2 x - 1)}{(u^2 + 1) \tan^{-1} u} dx$.
Since $du = \sin x \tan^2 x dx = \sin x (\sec^2 x - 1) dx$, we have $I = \int \frac{du}{(u^2 + 1) \tan^{-1} u}$.
Let $v = \tan^{-1} u$, then $dv = \frac{1}{u^2 + 1} du$.
Therefore, $I = \int \frac{1}{v} dv = \log|v| + c = \log(\tan^{-1}(\sec x + \cos x)) + c$.

(B) Let $I = \int \frac{\sqrt[n]{\text{cosec}^{2}x^n - 1}}{x^{(1-n)}} dx$.
Since $\text{cosec}^{2}x^n - 1 = \cot^{2}x^n$, the integral becomes $I = \int \frac{(\cot^{2}x^n)^{1/n}}{x^{1-n}} dx = \int \frac{(\cot x^n)^{2/n}}{x^{1-n}} dx$.
Let $u = \cot x^n$. Then $du = -\text{cosec}^{2}x^n \cdot n x^{n-1} dx$.
This substitution does not simplify directly. Let us use $u = x^n$, then $du = n x^{n-1} dx$, so $dx = \frac{du}{n x^{n-1}} = \frac{du}{n u^{(n-1)/n}}$.
$I = \int \frac{\sqrt[n]{\cot^2 u}}{x^{1-n}} \cdot \frac{du}{n x^{n-1}} = \int \frac{(\cot u)^{2/n}}{n} du$.
Actually, the standard form is $\int \frac{\cot x^n}{x^{1-n}} dx$. Let $u = x^n$, $du = n x^{n-1} dx$. Then $dx = \frac{du}{n x^{n-1}}$.
$I = \int \frac{\cot u}{x^{1-n}} \cdot \frac{du}{n x^{n-1}} = \frac{1}{n} \int \cot u du = \frac{1}{n} \log|\sin u| + c = \frac{1}{n} \log|\sin x^n| + c$.

(B) Let $I = \int \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1} dx$. Divide numerator and denominator by $x^{10}$:
$I = \int \frac{x^4(x^5 - x^{-5})}{x^{10} + 3 + x^{-10}} dx = \int \frac{x^9 - x^{-1}}{x^{10} + 3 + x^{-10}} dx$. This approach is complex. Instead, divide by $x^{10}$ inside the integral differently:
$I = \int \frac{x^{14}(1 - x^{-10})}{x^{20}(1 + 3x^{-10} + x^{-20})} dx = \int \frac{x^{-6}(1 - x^{-10})}{1 + 3x^{-10} + x^{-20}} dx$.
Let $u = x^5 + x^{-5}$. Then $du = (5x^4 - 5x^{-6}) dx = 5x^4(1 - x^{-10}) dx$.
$I = \frac{1}{5} \int \frac{du}{u^2 + 1} = \frac{1}{5} \tan^{-1}(u) + c = \frac{1}{5} \tan^{-1}(x^5 + x^{-5}) + c$.

(D) Let $t = x^2$, then $dt = 2x dx$, so $x dx = \frac{dt}{2}$.
$f(x) = \int \frac{1}{2(t+4)(t+9)} dt$.
Using partial fractions: $\frac{1}{(t+4)(t+9)} = \frac{1}{5} (\frac{1}{t+4} - \frac{1}{t+9})$.
$f(x) = \frac{1}{10} \int (\frac{1}{x^2+4} - \frac{1}{x^2+9}) dx$.
$f(x) = \frac{1}{10} [\frac{1}{2} \tan^{-1}(\frac{x}{2}) - \frac{1}{3} \tan^{-1}(\frac{x}{3})] + C$.
Wait, the integral is $\int \frac{x}{(x^2+4)(x^2+9)} dx = \frac{1}{2} \int \frac{dt}{(t+4)(t+9)} = \frac{1}{10} \ln|\frac{t+4}{t+9}| + C = \frac{1}{10} \ln|\frac{x^2+4}{x^2+9}| + C$.
Given $f(0) = \frac{1}{10} \ln(\frac{4}{9}) + C = \frac{1}{10} \ln((\frac{2}{3})^2) + C = \frac{1}{5} \ln(\frac{2}{3}) + C$.
Since $f(0) = \frac{1}{5} \ln(\frac{2}{3})$, we get $C = 0$.
Thus, $f(x) = \frac{1}{10} \ln(\frac{x^2+4}{x^2+9})$.
$f(1) = \frac{1}{10} \ln(\frac{1+4}{1+9}) = \frac{1}{10} \ln(\frac{5}{10}) = \frac{1}{10} \ln(\frac{1}{2}) = -\frac{1}{10} \ln(2)$.

(A) Let $I = \int (x^{21} + x^6 + x^3)(2x^{18} + 7x^3 + 14)^{1/3} dx$.
Factor out $x^{18}$ from the first term: $I = \int x^{18}(x^3 + x^{-12} + x^{-15})(2x^{18} + 7x^3 + 14)^{1/3} dx$.
Rewrite the integral as $I = \int x^{18} \cdot x^{18/3} (2 + 7x^{-15} + 14x^{-18})^{1/3} (x^3 + x^{-12} + x^{-15}) dx$.
$I = \int x^{24} (2 + 7x^{-15} + 14x^{-18})^{1/3} (x^3 + x^{-12} + x^{-15}) dx$.
Let $u = 2 + 7x^{-15} + 14x^{-18}$.
Then $du = (7(-15)x^{-16} + 14(-18)x^{-19}) dx = (-105x^{-16} - 252x^{-19}) dx = -21x^{-19}(5x^3 + 12) dx$.
Alternatively, factor out $x^{18}$ from the bracket: $I = \int x^{18} (x^3 + x^{-12} + x^{-15}) (x^{18})^{1/3} (2 + 7x^{-15} + 14x^{-18})^{1/3} dx = \int x^{24} (x^3 + x^{-12} + x^{-15}) (2 + 7x^{-15} + 14x^{-18})^{1/3} dx$.
Let $t = 2x^{18} + 7x^3 + 14$. Then $dt = (36x^{17} + 21x^2) dx = 3x^2(12x^{15} + 7) dx$.
$I = \int x^2(x^{19} + x^4 + x) (2x^{18} + 7x^3 + 14)^{1/3} dx$.
$I = \frac{1}{56} (2x^{18} + 7x^3 + 14)^{4/3} + c$.

(A) Step $1$: Complete the square for the quadratic expression $2x-x^2$.
$2x-x^2 = -(x^2-2x) = -(x^2-2x+1-1) = -( (x-1)^2 - 1 ) = 1 - (x-1)^2$.
Step $2$: Rewrite the integral using the completed square.
$\int \frac{1}{\sqrt{1-(x-1)^2}} dx$.
Step $3$: Use the standard integral formula $\int \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}(u) + c$.
Let $u = x-1$, then $du = dx$.
Therefore, $\int \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}(u) + c = \sin^{-1}(x-1) + c$.

(D) We have the integral $I = \int \frac{\sin^n x}{\cos^{n+2} x} dx$.
Rewrite the integrand as $I = \int \left( \frac{\sin x}{\cos x} \right)^n \cdot \frac{1}{\cos^2 x} dx$.
Since $\frac{\sin x}{\cos x} = \tan x$ and $\frac{1}{\cos^2 x} = \sec^2 x$, we get $I = \int \tan^n x \sec^2 x dx$.
Let $t = \tan x$, then $dt = \sec^2 x dx$.
Substituting these into the integral, we get $I = \int t^n dt$.
Using the power rule for integration, $I = \frac{t^{n+1}}{n+1} + C$.
Substituting $t = \tan x$ back, we get $I = \frac{\tan^{n+1} x}{n+1} + C$.