Indefinite Integration

Given integral: $I = \int \frac{(\sin ²x - \cos ²x)}{(\sin ²x \cos ²x) } dx$

Using the identity: $\sin ²x - \cos ²x = -\cos 2x$

The integral becomes: $I = \int \frac{(-\cos 2x) }{ (\sin ²x \cos ²x)} dx$

Using the identity:$\sin ²x \cos ²x = \left(\frac{1}{4}\right) \sin ²2x$

The integral transforms into: $I = -4\int \frac{\cos 2x}{\sin ²2x} dx$

Substituting t=sin2x, so that $dt = 2 \cos 2x dx$

Thus, $dx = \frac{dt}{(2 \cos 2x)}$

Substituting in the integral:

$$\begin{gathered} I = -4 \int \left(\frac{\cos 2x}{ t²}\right) \times \left(\frac{dt}{(2 \cos 2x)}\right) \\ = -4 \int \left(\frac{dt }{(2t²)}\right) \\ = -2 \int t⁻² dt \\ =\frac{2}{t}+c \end{gathered}$$

Using the identity we get: $\frac{1}{\sin x \cos x}= \tan x + cot x$+ c

Let $u=\log x\Rightarrow du=\frac{1}{x}dx$.

$\begin{array}{rcl}\int \frac{1}{x\left(\log x\right){}^2} \mathrm{d}x& =& \int \frac{1}{u^2}du\end{array}$

$\begin{array}{rcl}& =& \int u^{-2}du\end{array}$

$\begin{array}{rcl}& =& -\frac{1}{u}+C\end{array}$

$\begin{array}{rcl}& =& -\frac{1}{\log x}+C\end{array}$

Given Integral: $I=\int \frac{e^x(1+x)}{cos²(x.e^x)}dx$

Substituting $t=x{e}^x$

Differentiating both sides: $\frac{dt}{dx}=e^x(1+x)$

Rewriting the Integral in Terms of t: $I = \int \frac{dt }{\cos ² t}\Rightarrow \tan t$

$I=tan\left(x{e}^x\right)+C$

Given integral: $I = \int \frac{(\sin ²x - \cos ²x)}{(\sin ²x \cos ²x) } dx$

Using the identity: $\sin ²x - \cos ²x = -\cos 2x$

The integral becomes: $I = \int \frac{(-\cos 2x) }{ (\sin ²x \cos ²x)} dx$

Using the identity:$\sin ²x \cos ²x = \left(\frac{1}{4}\right) \sin ²2x$

The integral transforms into: $I = -4\int \frac{\cos 2x}{\sin ²2x} dx$

Substituting t=sin2x, so that $dt = 2 \cos 2x dx$

Thus, $dx = \frac{dt}{(2 \cos 2x)}$

Substituting in the integral:

$$\begin{gathered} I = -4 \int \left(\frac{\cos 2x}{ t²}\right) \times \left(\frac{dt}{(2 \cos 2x)}\right) \\ = -4 \int \left(\frac{dt }{(2t²)}\right) \\ = -2 \int t⁻² dt \\ =\frac{2}{t}+c \end{gathered}$$

Using the identity we get: $\frac{1}{\sin x \cos x}= \tan x + cot x$+ c

Let $u=\log x\Rightarrow du=\frac{1}{x}dx$.

$\begin{array}{rcl}\int \frac{1}{x\left(\log x\right){}^2} \mathrm{d}x& =& \int \frac{1}{u^2}du\end{array}$

$\begin{array}{rcl}& =& \int u^{-2}du\end{array}$

$\begin{array}{rcl}& =& -\frac{1}{u}+C\end{array}$

$\begin{array}{rcl}& =& -\frac{1}{\log x}+C\end{array}$

Given Integral: $I=\int \frac{e^x(1+x)}{cos²(x.e^x)}dx$

Substituting $t=x{e}^x$

Differentiating both sides: $\frac{dt}{dx}=e^x(1+x)$

Rewriting the Integral in Terms of t: $I = \int \frac{dt }{\cos ² t}\Rightarrow \tan t$

$I=tan\left(x{e}^x\right)+C$

$$\begin{gathered} f\left(x\right)=7{\tan }^{6}x(1+{\tan }^{2}x)-3{\tan }^{2}x(1+{\tan }^{2}x) \\ f\left(x\right)=se{c}^{2}x(7t{\mathrm{an}}^{6}x-3{\tan }^{2}x) \\ {I}_1=\int se{c}^{2}x(7t{\mathrm{an}}^{6}x-3{\tan }^{2}x) dx \\ let tanx=t \\ {I}_1=\frac{7t^7}{7}-3\frac{t^3}{3} \\ {I}_1=t{\mathrm{an}}^{7}x-{\tan }^{3}x+c_1 \\ {I}_2=\int xf\left(x\right)dx \\ {I}_2=\int xf\left(x\right)dx-\int 1.\int f\left(x\right)dx \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-\int (t{\mathrm{an}}^{7}x-{\tan }^{3}x)dx \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-\int {\tan }^{3}x({\tan }^{2}x-1)se{c}^{2}xdx \\ let \tan x=t ,se{c}^{2}xdx=dt \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-\int t^3(t^2-1)dt \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-\int (t^5-t^3)dt \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-(\frac{t^6}{6}-\frac{t^4}{4}) \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-(\frac{{\tan }^{6}x}{6}-\frac{{\tan }^{4}x}{4})+c_2 \\ now 7I_1+12I_2 \\ 7(t{\mathrm{an}}^{7}x-{\tan }^{3}x)+12\left(x\right(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-(\frac{\tan x^6}{6}-\frac{\tan x^4}{4}\left)\right)+c \\ \left(12x+7\right)\left({\tan }^{7}x-{\tan }^{3}x\right)+{\tan }^{4}x\left(3-2{\tan }^{2}x\right)+c \end{gathered}$$