Question

evaluate integration 0roπ/4 1/1+cot^3x
​

Answer 1

Topic :-

Integration

To Solve :-

$\sf{\displaystyle \int_{0}^{\pi/4}\dfrac{1}{1+cot^3x}dx}$

Solution :-

We will start by using property,

$\sf{\displaystyle \int_{a}^{b}f(x)dx=\displaystyle \int_{a}^{b}f(a+b-x)dx}$

Applying the property,

$\sf{\displaystyle\int_{0}^{\pi/4}\dfrac{1}{1+cot^3(0+\dfrac{\pi}{4}-x)}dx}$

Now, we can write

$\sf{cot(\dfrac{\pi}{4}-x)=\dfrac{cotxcot\dfrac{\pi}{4}+1}{cotx-cot\dfrac{\pi}{4}}}$

As,

$\sf {cot\dfrac{\pi}{4}=1}$

We can write,

$\sf{\dfrac{cotxcot\dfrac{\pi}{4}+1}{cotx-cot\dfrac{\pi}{4}}=\dfrac{cotx+1}{cotx-1}}$

From formula,

$\sf{cot(A-B)=\dfrac{cotBcotA+1}{cotB-cotA}}$

Now,

$\sf{\dfrac{cotx+1}{cotx-1}=\dfrac{\dfrac{1}{tanx}+1}{\dfrac{1}{tanx}-1}}$

$\sf{\dfrac{cotx+1}{cotx-1}=\dfrac{1+tanx}{1-tanx}}$

So, we can write

$\sf {\displaystyle\int_{0}^{\pi/4}\dfrac{1}{1+\left ( \dfrac{1+tanx}{1-tanx} \right )^3}dx}$

Now, we can write

$\sf{1+(\dfrac{1+tanx}{1-tanx})^3=\dfrac{(1-tanx)^3+(1+tanx)^3}{(1-tanx)^3}}$

From formula,

$\sf{(a+b)^3=a^3+b^3+3ab(a+b)}$

$\sf{(a-b)^3=a^3-b^3-3ab(a-b)}$

We can write,

$\sf{1+\left ( \dfrac{1+tanx}{1-tanx} \right )^3=\dfrac{6tan^2x+2}{1-tan^3x-3tanx+3tan^2x}}$

So, we can write,

$\sf {\displaystyle \int_{0}^{\pi/4}\dfrac{1-tan^3x-3tanx+3tan^2x}{6tan^2x+2}dx}$

$\sf {\displaystyle \int_{0}^{\pi/4}\dfrac{3tan^2x+1-tan^3x-3tanx}{2(3tan^2x+1)}dx}$

$\sf{\dfrac{1}{2}\sf {\displaystyle \int_{0}^{\pi/4}1-(\dfrac{tan^3x+3tanx}{3tan^2x+1})dx}}$

$\sf{\dfrac{1}{2}\displaystyle \int_{0}^{\pi/4}dx-\dfrac{1}{2}\displaystyle \int_{0}^{\pi/4}(\dfrac{sec^2x(tan^3x+3tanx)}{(1+tan^2x)(3tan^2x+1)})dx}$

Substitute, u = tanx

$\sf{du=sec^2xdx}$

Limit changes to :

$\sf {u=tan\dfrac{\pi}{4}=1}$

u = tan(0) = 0

$\sf{\dfrac{1}{2}\left [ x \right ]^\frac{\pi}{4}_0 -\dfrac{1}{2}\displaystyle \int_{0}^{1}\dfrac{u^3+3u}{(u^2+1)(3u^2+1)}du}$

$\sf{\dfrac{\pi}{8}-\dfrac{1}{2}\displaystyle \int_{0}^{1}\dfrac{4u}{3u^2+1}-\dfrac{u}{u^2+1}du}$

Upon solving we get,

$\sf{\dfrac{\pi}{8}-\dfrac{1}{2}\left [ \dfrac{4ln(3u^2+1)}{6}-\dfrac{ln(u^2+1)}{2} \right ]^1_0}$

$\sf{\dfrac{\pi}{8}-\dfrac{1}{2}\left [ \dfrac{4ln(3(1)^2+1)}{6}-\dfrac{ln((1)^2+1)}{2} -\left ( \dfrac{4ln(3(0)^2+1)}{6}-\dfrac{ln((0)^2+1)}{2} \right ) \right ]}$

$\sf{\dfrac{\pi}{8}-\dfrac{1}{2}\left [ \dfrac{4ln(4)}{6}-\dfrac{ln(2)}{2} -\left ( \dfrac{4ln(1)}{6}-\dfrac{ln(1)}{2} \right ) \right ]}$

As we know, ln(1) = 0.

So,

$\sf{\dfrac{\pi}{8}-\dfrac{1}{2}\left [ \dfrac{8ln(2)}{6}-\dfrac{ln(2)}{2} \right ]}$

$\sf{\dfrac{\pi}{8}-\dfrac{1}{2}ln(2)\left [ \dfrac{4}{3}-\dfrac{1}{2} \right ]}$

$\sf{\dfrac{\pi}{8}-\dfrac{1}{2}ln(2)\left [ \dfrac{8-3}{6} \right ]}$

$\sf{\dfrac{\pi}{8}-\dfrac{5}{12}ln(2)}$

Answer :-

$\sf{So,\:answer\:is,\:\:\dfrac{\pi}{8}-\dfrac{5}{12}ln(2)}$