Question

Integration


Calculus​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: I \: = \displaystyle\int_0^1 \tt \: {x}^{6} \sqrt{1 - {x}^{2} } \: dx$

We use here method of Substitution,

$\red{\rm :\longmapsto\:Put \: x = siny} \\ \red{\rm :\longmapsto\:dx = cosy \: dy}$

☆ Now, we have to change the limits too

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf \dfrac{\pi}{2} \end{array}} \\ \end{gathered}$

☆ So, given integral can be rewritten as

$\rm :\longmapsto\: I \: = \displaystyle\int_0^{\dfrac{\pi}{2}} \tt \: {sin}^{6}y\sqrt{1 - {sin}^{2}y} \: cosy \: dy$

$\rm :\longmapsto\: I \: = \displaystyle\int_0^{\dfrac{\pi}{2}} \tt \: {sin}^{6}y\sqrt{{cos}^{2}y} \: cosy \: dy$

$\rm :\longmapsto\: I \: = \displaystyle\int_0^{\dfrac{\pi}{2}} \tt \: {sin}^{6}y \: {cos}^{2}y\: \: dy$

☆ Using Walli's Formula,

$\rm :\longmapsto\:I \: = \dfrac{(5 \times 3 \times 1)(1)}{8 \times 6 \times 4 \times 2} \times \dfrac{\pi}{2}$

$\bf :\longmapsto\:I \: = \dfrac{5\pi}{256}$

Additional Information :-

Walli's Formula:-

1. If n is odd

$\displaystyle\int_0^{\dfrac{\pi}{2}} \tt \: {sin}^{n}x \: dx = \dfrac{n - 1}{n} .\dfrac{n - 3}{n - 2} .\dfrac{n - 5}{n - 4} . - - - \dfrac{3}{2}$

2. If n is even,

$\displaystyle\int_0^{\dfrac{\pi}{2}} \tt \: {sin}^{n}x \: dx = \dfrac{n - 1}{n} .\dfrac{n - 3}{n - 2} .\dfrac{n - 5}{n - 4} . - - - \dfrac{1}{2}.\dfrac{\pi}{2}$

3. If n is even,

$\displaystyle\int_0^{\dfrac{\pi}{2}} \tt \: {cos}^{n}x \: dx = \dfrac{n - 1}{n} .\dfrac{n - 3}{n - 2} .\dfrac{n - 5}{n - 4} . - - - \dfrac{1}{2}.\dfrac{\pi}{2}$

4. If n is odd,

$\displaystyle\int_0^{\dfrac{\pi}{2}} \tt \: {cos}^{n}x \: dx = \dfrac{n - 1}{n} .\dfrac{n - 3}{n - 2} .\dfrac{n - 5}{n - 4} . - - - \dfrac{3}{2}$