Question

Q -1) Evaluate :-


${ \bf{ \int \frac{1}{ \cos {}^{4}x \: + \: \sin {}^{4}x } dx}}$
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Answer 1

We're given to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{1}{\cos^4x+\sin^4x}\ dx$

Dividing numerator and denominator by $\cos^4x,$

$\displaystyle\longrightarrow I=\int\dfrac{\left(\dfrac{1}{\cos^4x}\right)}{\left(\dfrac{\cos^4x+\sin^4x}{\cos^4x}\right)}\ dx$

or,

$\displaystyle\longrightarrow I=\int\dfrac{\sec^4x}{1+\tan^4x}\ dx$

$\displaystyle\longrightarrow I=\int\dfrac{\sec^2x\cdot\sec^2x}{1+\tan^4x}\ dx$

Since $\sec^2x=1+\tan^2x,$

$\displaystyle\longrightarrow I=\int\dfrac{1+\tan^2x}{1+(\tan^2x)^2}\cdot\sec^2x\,dx$

$\displaystyle\longrightarrow I=\int\dfrac{1+\tan^2x}{(1+\tan^2x)^2-2\tan^2x}\cdot\sec^2x\,dx\quad\quad\dots(1)$

Substitute,

$\longrightarrow u=\tan x$

$\longrightarrow du=\sec^2x\ dx$

Then (1) becomes,

$\displaystyle\longrightarrow I=\int\dfrac{1+u^2}{(1+u^2)^2-2u^2}\ du$

$\displaystyle\longrightarrow I=\int\dfrac{1+u^2}{(1+u^2)^2-\left(u\sqrt2\right)^2}\ du$

Factorising the denominator,

$\displaystyle\longrightarrow I=\int\dfrac{1+u^2}{(1+u\sqrt2+u^2)(1-u\sqrt2+u^2)}\ du$

or,

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{\left(1+u\sqrt2+u^2\right)+\left(1-u\sqrt2+u^2\right)}{(1+u\sqrt2+u^2)(1-u\sqrt2+u^2)}\ du$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\left(\dfrac{1}{u^2+u\sqrt2+1}+\dfrac{1}{u^2-u\sqrt2+1}\right)\ du$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{1}{u^2+u\sqrt2+1}\ du+\dfrac{1}{2}\int\dfrac{1}{u^2-u\sqrt2+1}\ du$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\dfrac{1}{\left(u+\dfrac{1}{\sqrt2}\right)^2+\left(\dfrac{1}{\sqrt2}\right)^2}\ du+\dfrac{1}{2}\int\dfrac{1}{\left(u-\dfrac{1}{\sqrt2}\right)^2+\left(\dfrac{1}{\sqrt2}\right)^2}\ du$

We can directly apply this formula,

• $\displaystyle\int\dfrac{1}{(ax+b)^2+c^2}\ dx=\dfrac{1}{ac}\tan^{-1}\left(\dfrac{ax+b}{c}\right)+C$

Consider the first integral (ignoring the coefficient),

$\displaystyle\longrightarrow I_1=\int\dfrac{1}{\left(u+\dfrac{1}{\sqrt2}\right)^2+\left(\dfrac{1}{\sqrt2}\right)^2}\ du$

On comparing with the formula, we get,

• $a=1$

• $x=u$

• $b=\dfrac{1}{\sqrt2}$

• $c=\dfrac{1}{\sqrt2}$

Thus,

$\displaystyle\longrightarrow I_1=\dfrac{1}{1\cdot\dfrac{1}{\sqrt2}}\tan^{-1}\left(\dfrac{u+\dfrac{1}{\sqrt2}}{\dfrac{1}{\sqrt2}}\right)+C$

$\displaystyle\longrightarrow I_1=\sqrt2\tan^{-1}\left(u\sqrt2+1\right)+C$

Similarly, the second integral,

$\displaystyle\longrightarrow I_2=\int\dfrac{1}{\left(u-\dfrac{1}{\sqrt2}\right)^2+\left(\dfrac{1}{\sqrt2}\right)^2}\ du$

will be,

$\displaystyle\longrightarrow I_2=\sqrt2\tan^{-1}\left(u\sqrt2-1\right)+C$

Hence,

$\displaystyle\longrightarrow I=\dfrac{1}{2}\left[\sqrt2\tan^{-1}\left(u\sqrt2+1\right)+\sqrt2\tan^{-1}\left(u\sqrt2-1\right)\right]+C$

$\displaystyle\longrightarrow I=\dfrac{1}{\sqrt2}\left[\tan^{-1}\left(u\sqrt2+1\right)+\tan^{-1}\left(u\sqrt2-1\right)\right]+C$

$\displaystyle\longrightarrow I=\dfrac{1}{\sqrt2}\tan^{-1}\left(\dfrac{u\sqrt2+1+u\sqrt2-1}{1-\left(u\sqrt2+1\right)\left(u\sqrt2-1\right)}\right)$

$\displaystyle\longrightarrow I=\dfrac{1}{\sqrt2}\tan^{-1}\left(\dfrac{2u\sqrt2}{2-2u^2}\right)+C$

$\displaystyle\longrightarrow I=\dfrac{1}{\sqrt2}\tan^{-1}\left(\dfrac{u\sqrt2}{1-u^2}\right)+C$

Undoing substitution $u=\tan x,$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{\sqrt2}\tan^{-1}\left(\dfrac{\sqrt2\tan x}{1-\tan^2x}\right)+C}}$