Substitution

Let $I_n=\displaystyle\int\tan^nx\,dx$, $(n>1)$. If $I_4+I_6=a\tan^5x+bx^5+C$, where C is a constant of integration, then the ordered pair (a, b) is equal to :

The integral $\displaystyle\int_{\pi/12}^{\pi/4}\frac{8\cos2x}{(\tan x+\cot x)^3}\,dx$ equals :

If $f\left(\dfrac{3x-4}{3x+4}\right)=x+2$, $x\ne-\dfrac43$, and $\displaystyle\int f(x)\,dx=A\log|1-x|+Bx+C$, then the ordered pair (A, B) is equal to :

(where C is a constant of integration)

The integral $\displaystyle\int\frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,dx$ is equal to :

where C is an arbitrary constant.

If $\displaystyle\int\frac{dx}{\cos^3x\sqrt{2\sin2x}}=(\tan x)^A+C(\tan x)^B+k$, where k is a constant of integration, then A + B + C equals :

The integral $\displaystyle\int\frac{dx}{(1+\sqrt x)\sqrt{x-x^2}}$ is equal to :

(where C is a constant of integration.)

If
$$\alpha=\int_0^1\left(e^{9x+3\tan^{-1}x}\right)\left(\frac{12+9x^2}{1+x^2}\right)dx$$
where $\tan^{-1}x$ takes only principal values, then the value of $\left(\log_e|1+\alpha|-\dfrac{3\pi}4\right)$ is

Let $f(x)=7\tan^8x+7\tan^6x-3\tan^4x-3\tan^2x$ for all $x\in\left(-\dfrac\pi2,\dfrac\pi2\right)$. Then the correct expression(s) is(are)

The integral $\displaystyle\int\frac{dx}{x^2(x^4+1)^{3/4}}$ equals :

The integral $\displaystyle\int\frac{dx}{(x+1)^{\frac34}(x-2)^{\frac54}}$ is equal to :

If $\displaystyle\int\frac{\log\left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}}dt=\frac12(g(t))^2+C$, where C is a constant, then g(2) is equal to :

The following integral
$$\int_{\pi/4}^{\pi/2}(2\operatorname{cosec}x)^{17}\,dx$$
is equal to

The value of $g\left(\frac12\right)$ is

The integral $\displaystyle\int_0^{1/2}\frac{\ln(1+2x)}{1+4x^2}\,dx$, equals :

The integral $\displaystyle\int\frac{\sin^2x\cos^2x}{(\sin^3x+\cos^3x)^2}dx$ is equal to :

If $m$ is a non - zero number and
$$\int\frac{x^{5m-1}+2x^{4m-1}}{(x^{2m}+x^m+1)^3}dx=f(x)+c,$$
then $f(x)$ is :

The integral $\displaystyle\int\frac{\sec^2x}{(\sec x+\tan x)^{9/2}}\,dx$ equals (for some arbitrary constant $K$)

The value of
$$\int_{\sqrt{\ln2}}^{\sqrt{\ln3}}\frac{x\sin x^2}{\sin x^2+\sin\left(\ln6-x^2\right)}\,dx$$
is

The area of the region between the curves $y=\sqrt{\dfrac{1+\sin x}{\cos x}}$ and $y=\sqrt{\dfrac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\frac\pi4$ is

Let
$$I=\int\frac{e^x}{e^{4x}+e^{2x}+1}dx,\qquad J=\int\frac{e^{-x}}{e^{-4x}+e^{-2x}+1}dx.$$
Then, for an arbitrary constant $C$, the value of $J-I$ equals