Indefinite integratiom

for first one its simple convert sinx + cos x to √2sin(pi/4+x)
then let pi/4+x=z and continue..........

the second one u can evn do it by takin the whole Denominator as some t and thn apply the formulae....but the options arent matching! :

dude u cn take the whole denmtor as a t2 it wil smplify your wrk
and for 5th sum i think u mst cnvrt 1/cos²x as sec² and proceed by ∫u.vdx

wel its one and the same..!!

Ill try the 5th one tht way..!!

mch btr......
hpe u gt the answer

\hspace{-16}(4)\;\; \int \left\{1+\cot (x+\alpha).\cot(x-\alpha)\right\}dx\\\\ $Now Using Trigonometric Identity::\\\\ $\cot\left\{(x+\alpha)-(x-\alpha)\right\}=\frac{\cot (x+\alpha).\cot(x-\alpha)+1}{\cot (x-\alpha)-\cot (x+\alpha)}$\\\\\\ So $\boxed{1+\cot (x+\alpha).\cot(x-\alpha)=\cot(2\alpha).\left\{\cot (x-\alpha)-\cot (x+\alpha)\right\}}$\\\\\\ So Integral Convert into $\cot(2\alpha).\int \left\{\cot (x-\alpha)-\cot (x+\alpha)\right\}dx$\\\\\\ $=\cot(2\alpha).\int \cot (x-\alpha)dx -\cot(2\alpha).\int \cot (x+\alpha)dx$\\\\\\ $=\cot(2\alpha).\ln\mid \sin(x-\alpha) \mid - \cot(2\alpha).\ln\mid \sin(x+\alpha) \mid+C$\\\\\\ $=\cot(2\alpha).\ln \mid \frac{\sin(x-\alpha)}{\sin(x+\alpha)} \mid +C$

\hspace{-16}(3)::\int\frac{\left(\sec^7 x-7\right)}{\sin^2 x}dx=\int\frac{\sec^7 x}{\sin^2 x}-7.\int\frac{1}{\sin^2 x}dx\\\\\\ =\int \underbrace{\sec^7 x}_{1}.\underbrace{\csc^2 x}_{2}dx-7.\int\csc^2 xdx$\\\\\\ Apply integration By parts for First, We Get\\\\ $=\sec^7 x.(-\cot x)-7.\int\sec^6 x.\sec x.\tan x.(-\cot x)dx-7(-\cot x)+C $\\\\\\ $=-\sec^7 x.\cot x+7.\int \sec^7 xdx+7\cot x+C$\\\\ Now We Calculate value of $\int\sec^7 x dx$\\\\\\ So First We Calculate for $\int\sec^n x dx$\\\\\\ Let $I_{n}=\int\sec^n x dx=\int\sec^{n-2}x.\sec^2 xdx$\\\\\\ $I_{n}=\sec^{n-2}x.\tan x-(n-2).\int \sec^{n-3}x.\sec x.\tan x.\tan xdx$\\\\\\ $I_{n}=\sec^{n-2}x.\tan x-(n-2)\int \sec^{n-2}.(\sec^2 x-1)dx$\\\\\\ $I_{n}=\sec^{n-2}x.\tan x-(n-2).\int \sec^n xdx+(n-2).\int\sec^{n-2}xdx$\\\\\\ $I_{n}=\sec^{n-2}x.\tan x-(n-2).I_{n}+(n-2).I_{n-2}$\\\\\\ $I_{n}=\frac{\sec^{n-2}.\tan x}{n-1}+\frac{n-2}{n-1}.I_{n-2}$

\hspace{-16} $Replace $n\rightarrow (n-2)\;,$ We Get\\\\ $I_{n-2}=\frac{\sec^{n-4}.\tan x}{n-3}+\frac{n-4}{n-3}.I_{n-4}$\\\\ Again Replace $n\rightarrow (n-2)\;,$ We Get\\\\ $I_{n-4}=\frac{\sec^{n-6}.\tan x}{n-5}+\frac{n-6}{n-5}.I_{n-6}$\\\\ Now Put This value in $I_{n-2}\;,$ We Get\\\\ $I_{n-2}=\frac{\sec^{n-4}.\tan x}{n-3}+\frac{n-4}{n-3}.\left\{\frac{\sec^{n-6}.\tan x}{n-5}+\frac{n-6}{n-5}.I_{n-6}\right\}$\\\\ so $I_{n-2}=\frac{\sec^{n-4}.\tan x}{n-3}+\frac{(n-4)}{(n-3).(n-5)}.\sec^{n-6}.\tan x+\frac{(n-4).(n-6)}{(n-3).(n-5)}.I_{n-6}$\\\\\\ So put value of $I_{n-2}$ in $I_{n}\;,$ We Get\\\\ $I_{n}=\frac{\sec^{n-2}.\tan x}{n-1}+\frac{(n-2)}{(n-1).(n-3)}.\sec^{n-4}.\tan x+\frac{(n-2).(n-4)}{(n-1).(n-3).(n-5)}.\sec^{n-6}.\tan x+\\\\\frac{(n-2).(n-4).(n-6)}{(n-1).(n-3).(n-5)}.I_{n-6}$\\\\\\ Put $n=7\;,$ We Get\\\\ $I_{7}=\int \sec^7 xdx=\frac{1}{6}.\sec^5 x.\tan x+\frac{5}{24}.\sec^3 x.\tan x+\frac{5}{16}.\sec x.\tan x+\\\\\frac{5}{16}\ln\mid \sec x+\tan x \mid+C$

Where Ei = exponential Integral

I Think not in terms of elementry function.

5) ∫log(cosx) sec²x dx
just use byparts to get answer as,
tanx.log(cosx) + tanx - x

for the last question by shubham,
take log(x + ....) as t so the answer is 4pi *t²2

u can also do the first Q by taking tanx2 = t.