Question

evaluate integration limit0 to π\2 sin²xcos³x x dx class 11​

Answer 1

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

Let assume that

$\rm \longrightarrow \: I = \displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {sin}^{2}x \: {cos}^{3}x \: dx$

can be rewritten as

$\rm \longrightarrow \: I = \displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {sin}^{2}x \: {cos}^{2}x \: cosx \: dx$

$\rm \longrightarrow \: I = \displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {sin}^{2}x \:(1 - {sin}^{2}x )\: cosx \: dx$

Now, we use substitution method to solve this integral.

$\red{\rm :\longmapsto\:Put \: sinx = y} \\ \red{\rm :\longmapsto\:cosx \: dx = dy}$

When we use substitution method, we have to change the limits.

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

So, given integral reduced to

$\rm \longrightarrow \: I = \displaystyle \int_{0}^{1} \rm \: {y}^{2}\:(1 - {y}^{2})\: \: dy$

$\rm \longrightarrow \: I = \displaystyle \int_{0}^{1} \rm \: ({y}^{2}\: - {y}^{4})\: \: dy$

$\rm \longrightarrow \: I = \bigg(\dfrac{ {y}^{3} }{3} - \dfrac{ {y}^{5} }{5} \bigg)_0^1$

$\rm \longrightarrow \: I = \bigg(\dfrac{1 }{3} - \dfrac{1}{5} \bigg)$

$\rm \longrightarrow \: I = \bigg(\dfrac{5 - 3}{15}\bigg)$

$\rm \longrightarrow \: I = \dfrac{2}{15}$

Additional Information :-

Properties of Integration :-

$\displaystyle\sf (1) \int\limits^b_a {f(x)} \, dx = \int\limits^b_a {f(t)}dt$

$\displaystyle\sf (2) \int\limits^b_a {f(x)} \, dx = - \: \int\limits^a_b {f(x)}dx$

$\sf (3) \displaystyle\int\limits^b_a {f(x)} \, dx = \int\limits^c_a {f(x)} \, dx + \int\limits^b_c {f(x)} \, dx \ where\ \ a < c < b$

$\sf (4) \displaystyle \int\limits^a_0 {f(x)} \, dx =\int\limits^a_0 {f(a - x)}$

$\sf (5) \displaystyle\int\limits^b_a {f(x)} \, dx = \int\limits^b_a{f(a + b - x)}$