Trigonometric Integrals

Using $\sin^2\theta = \frac{1-\cos 2\theta}{2}$:
$\int \frac{1 – \cos(4x+10)}{2} dx = \frac{1}{2}\left(x – \frac{\sin(4x+10)}{4}\right) + C$.

Multiply/Divide by 2: $\frac{1}{2} \int (\sin 7x – \sin x) dx$.
$= \frac{1}{2} \left( -\frac{\cos 7x}{7} + \cos x \right) + C$.

Group first two terms: $\frac{1}{2} \int (\cos 6x + \cos 2x)\cos 6x dx$.
$= \frac{1}{2} \int (\cos^2 6x + \cos 2x \cos 6x) dx$.
$= \frac{1}{2} \int (\frac{1+\cos 12x}{2} + \frac{\cos 8x + \cos 4x}{2}) dx$.
$= \frac{1}{4} [x + \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + \frac{\sin 4x}{4}] + C$.

Use $\sin 3\theta = 3\sin\theta – 4\sin^3\theta \implies \sin^3\theta = \frac{3\sin\theta – \sin 3\theta}{4}$.
$\int \frac{3\sin(2x+1) – \sin(6x+3)}{4} dx$.
$= -\frac{3}{8}\cos(2x+1) + \frac{1}{24}\cos(6x+3) + C$.

Let $\sin x = t \implies \cos x dx = dt$.
$\int t^3 \cos^2 x dt = \int t^3 (1-t^2) dt = \int (t^3 – t^5) dt$.
$= \frac{t^4}{4} – \frac{t^6}{6} + C = \frac{1}{4}\sin^4 x – \frac{1}{6}\sin^6 x + C$.

Similar to Q3. Combine $\sin x \sin 2x$ first.
$= \frac{1}{4} \left[ \frac{\cos 2x}{2} – \frac{\cos 4x}{4} – \frac{\cos 6x}{6} \right] \dots \text{(after expanding)}$.
Result: $\frac{1}{4} [ \frac{1}{12}\cos 6x – \frac{1}{4}\cos 4x – \frac{1}{2}\cos 2x ] + C$.

$= \int \frac{2\sin^2(x/2)}{2\cos^2(x/2)} dx = \int \tan^2(x/2) dx$.
$= \int (\sec^2(x/2) – 1) dx = 2\tan(x/2) – x + C$.

$= \int \frac{1+\cos x – 1}{1+\cos x} dx = \int (1 – \frac{1}{2\cos^2(x/2)}) dx$.
$= \int (1 – \frac{1}{2}\sec^2(x/2)) dx = x – \tan(x/2) + C$.

$(\sin^2 x)^2 = (\frac{1-\cos 2x}{2})^2 = \frac{1}{4}(1 – 2\cos 2x + \cos^2 2x)$.
Substitute $\cos^2 2x = \frac{1+\cos 4x}{2}$. Integrate term by term.
Result: $\frac{3x}{8} – \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$.

Numerator: $(2\cos^2 x – 1) – (2\cos^2\alpha – 1) = 2(\cos^2 x – \cos^2\alpha)$.
$= 2(\cos x – \cos\alpha)(\cos x + \cos\alpha)$.
Cancel denominator: $\int 2(\cos x + \cos\alpha) dx = 2\sin x + 2x\cos\alpha + C$.

Denominator: $\cos^2 x + \sin^2 x + 2\sin x \cos x = (\cos x + \sin x)^2$.
Let $\cos x + \sin x = t \implies (-\sin x + \cos x)dx = dt$.
$\int \frac{dt}{t^2} = -\frac{1}{t} = -\frac{1}{\cos x + \sin x} + C$.

$= \int (\tan^2 2x)(\sec 2x \tan 2x) dx = \int (\sec^2 2x – 1)(\sec 2x \tan 2x) dx$.
Let $\sec 2x = t \implies 2\sec 2x \tan 2x dx = dt$.
$\frac{1}{2} \int (t^2 – 1) dt = \frac{1}{6}\sec^3 2x – \frac{1}{2}\sec 2x + C$.

Numerator $1 = \sin^2 x + \cos^2 x$.
$\int (\frac{\sin^2 x}{\sin x \cos^3 x} + \frac{\cos^2 x}{\sin x \cos^3 x}) dx = \int (\tan x \sec^2 x + \frac{1}{\sin x \cos x}) dx$.
Second term: multiply num/den by $\cos x \to \int \frac{\sec^2 x}{\tan x} dx$.
Result: $\frac{1}{2}\tan^2 x + \log|\tan x| + C$.

Multiply and divide by $\sin(a-b)$. Note: $a-b = (x-b) – (x-a)$.
$\frac{1}{\sin(a-b)} \int \frac{\sin((x-b)-(x-a))}{\cos(x-a)\cos(x-b)} dx$.
Expand sin(A-B) and split.
Result: $\frac{1}{\sin(a-b)} [\log|\cos(x-b)| – \log|\cos(x-a)|] + C$.

Split: $\int (\frac{1}{\cos^2 x} – \frac{1}{\sin^2 x}) dx = \int (\sec^2 x – \text{cosec}^2 x) dx$.
$= \tan x – (-\cot x) + C = \tan x + \cot x + C$.

Let $e^x x = t \implies (e^x \cdot 1 + x e^x) dx = dt \implies e^x(1+x)dx = dt$.
$\int \frac{dt}{\cos^2 t} = \int \sec^2 t dt = \tan t + C = \tan(xe^x) + C$.