Question

integrate x sin x^2 dx limits 0 to root pi​

Answer 1

Given Question :-

Evaluate:-

$\sf \ \displaystyle\int^{ \sqrt{\pi} } _{0} \tt \: x \: sin(x^{2})dx$

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

Formula used :-

$\boxed{\bf\: \int \: sinx \: dx = - \: cosx}$

$\boxed{\bf \: cos\pi = - 1}$

$\boxed{\bf \: cos0= 1}$

Let's solve the problem now!!

Given that

$\sf \ \displaystyle\int^{ \sqrt{\pi} } _{0} \tt \: x \: sin(x^{2})dx$

$\rm \: Let \:I \: = \: \sf \ \displaystyle\int^{ \sqrt{\pi} } _{0} \tt \: x \: sin(x^{2})dx$

As we know that,

By using substitution method, we get.

• ⇒ Put x² = y.

Differentiate w.r.t x, we get.

• ⇒ 2x dx = dy.

• ⇒ (x)dx = dy/2.

As we know that,

• In definite integration, if we apply substitution method, then limit will also change, so, we get,

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

Hence,

Given integral can be rewritten as

$\rm :\longmapsto\: I \: = \: \displaystyle\int^{\pi} _{0} \tt \: \: siny \: \dfrac{dy}{2}$

$\rm :\longmapsto\: I \: = \dfrac{1}{2} \: \displaystyle\int^{\pi} _{0} \tt \: \: siny \: dy$

$\rm :\longmapsto\: I \: = - \dfrac{1}{2} \: \bigg(cosy\bigg)^{\pi} _{0} \tt \:$

$\rm :\longmapsto\: I \: = - \dfrac{1}{2} \: \bigg(cos\pi \: - \: cos0\bigg)$

$\rm :\longmapsto\: I \: = - \dfrac{1}{2} \: \bigg( - 1\: - \: 1\bigg)$

$\rm :\longmapsto\: I \: = - \dfrac{1}{2} \: \bigg( - \: 2\bigg)$

$\rm :\longmapsto\: I \: = 1$

Hence,

$\pink{\underbrace{ \purple{\boxed{\bf \:\sf \ \displaystyle\int^{ \sqrt{\pi} } _{0} \tt \: x \: sin(x^{2})dx \: = \: 1}}}}$

Additional Information :-

Properties of definite integration :-

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

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

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

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

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