Maths

Kindly go through the solution

$\begin{array}{l}\begin{array}{l}Let{x}^3=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}3x^{2}dx=dt\hfill \end{array}\\ I=\int \frac{x^2}{1-x^6}dx=\frac{1}{3}\int \frac{dt}{1-t^2}\\ =\frac{1}{3}\left[\frac{1}{2}log\left|\frac{1+t}{1-t}\right|\right]+C\\ =\frac{1}{6}log\left|\frac{1+x^3}{1-x^3}\right|+C\end{array}$

$\begin{array}{l}Le\mathrm{t\surd 2}{x}^2=t\\ \Rightarrow \mathrm{2\surd 2}xdx=dt\\ I=\int \frac{3x}{1+2x^4}dx\\ =\frac{3}{\mathrm{2\surd 2}}\int \frac{dt}{1+t^2}\\ =\frac{3}{\mathrm{2\surd 2}}\left[ta{n}^{-1}t\right]+C\\ =\frac{3}{\mathrm{2\surd 2}}\left[ta{n}^{-1}(\surd 2x^2)\right]+C\end{array}$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

$\begin{array}{l}\begin{array}{l}Let{x}^3=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}3x^{2}dx=dt\hfill \end{array}\\ I=\int \frac{3x^2}{x^6+1}dx=\int \frac{dt}{t^2+1}\\ \begin{array}{ll}= ta{n}^{–1}t+C\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= ta{n}^{–1}\left(x^3\right) +C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Let{e}^{2}x=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}{e}^{2}x+ e^{x}1dx=dt\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}{e}^x\left(x+1\right)dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{e^x(1+x)}{co{s}^2\left(e^{2}x\right)}}dx={\displaystyle \int \frac{dt}{co{s}^{2}t}}\\ \begin{array}{l}={\displaystyle \int se{c}^2}t dt = tan \left(e^x,x\right) + C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\therefore The correct answer is \left(B\right).\hfill \end{array}\end{array}$

$\begin{array}{l}={\displaystyle \int \frac{si{n}^{2}x-co{s}^{2}x}{si{n}^{2}co{s}^{2}x}}dx\\ \begin{array}{l}={\displaystyle \int (se{c}^{2}x-cose{c}^{2}x)}dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= tanx+ cotx+ C.\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Therefore, the correct answer is \left(A\right).\hfill \end{array}\end{array}$

$\begin{array}{l}\frac{1}{cos(x-a)cos(x-b)}=\frac{1}{sin(a-b)}*\left[\frac{sin(a-b)}{cos(x-a)cos(x-b)}\right]\\ =\frac{1}{sin(a-b)}\left[\frac{sin\{(x-b)-(x-a)\}}{cos(x-a)cos(x-b)}\right]\end{array}$

$\begin{array}{l}=\frac{1}{sin(a-b)}[sin(x-b)cos(x-a)-cos(x-b)\cdot sin(x-a)cos(x-a)cos(x-b)\\ =\frac{1}{sin(a-b)}[tan(x-b)-tan(x-a)]\\ =\frac{1}{sin(a-b)}{\displaystyle \int tan}(x-b)-tan\left(x-a\right)]dx\\ =\frac{1}{sin(a-b)}[-log|cos(x-b)|+log|cos(x-a)?]\\ =\frac{1}{sin(a-b)}\left[log\left|\frac{cos(x-a)}{cos(x-b)}\right|\right]+\; C\end{array}$