Integral of Trigonometric Functions

Evaluate : $\int \sqrt{\tan \theta } d\theta$ [Take $\sqrt{\tan \theta }=t$]

$\int \frac{\sin x}{1+\sin x}dx$

$\int \sqrt[3]{\frac{{\sin }^{2}x}{{\cos }^{14}x}}dx$

$\int \left\{1+\tan x.\tan (x+\alpha )\right\}dx=$

Find the area of the region bounded below by the curve $y=\frac{1}{2},$ to the left by the curve $y=\cos x$ and above by the curve $y=\sin x$, given that $x\in \left(0,2\mathrm{\pi }\right)$

If $\int \frac{1+\cos 8x}{\tan 2x-\cot 2x}dx=f\left(x\right)·\cos \left(g\left(x\right)\right)+c$, then $f\left(\frac{1}{4}\right)+g\left(\frac{1}{4}\right)=$

$\int {\sin }^{3}xdx+\int {\cos }^{2}x\sin xdx=$

$\int \left({\tan }^2\left(2x\right)-{\cot }^2\left(2x\right)\right)dx=$

If ${\int }_0^{\mathrm{\pi }}\left[{\cos }^2\left(\frac{3\mathrm{\pi }}{8}-\frac{x}{4}\right)-{\cos }^2\left(\frac{11\mathrm{\pi }}{8}+\frac{x}{4}\right)\right]dx$ equal $\sqrt{k}.$ Then find the value of $k.$

Evaluate ${\int }_0^{\frac{\mathrm{\pi }}{2}}\left|\cos x-\sin x\right|dx$

If $\int \frac{\cos x+x}{1+\sin x}dx=f\left(x\right)+\int \frac{3\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin {\displaystyle \frac{x}{2}}}dx+c$, then $f\left(x\right)=$

Integrate  w.r.t.  x :

(i) $\frac{{(1+\log x)}^3}{x}$

Integrate  w.r.t.  x :

(i) $\sin x {\cos }^{4}x$

Integrate  w.r.t.  x :

(i) $\sqrt{1-\sin x}$

Integrate  w.r.t.  x :

(i) ${\cos }^{3}x$

Integrate  w.r.t.  x :

(i) ${\left(\sin x-\cos x\right)}^2$

Integrate  w.r.t.  x :

(i) ${\cos }^2\left(\frac{x}{2}\right)$

Integrate  w.r.t.  x :

(i) $\frac{1}{(1+\sin x)}$

If $\int \frac{1+\cos \left(4x\right)}{\cot \left(x\right)-\tan \left(x\right)}dx=A\cos \left(4x\right)+B$, then $A=$