Question

A revision question for jee advanced.

Evaluate the given integral.

$\displaystyle\int_{\sqrt{ln2}}^\sqrt{ln3}} \dfrac{xsin(x^2)}{sin(x^2)+sin(ln6-x^2)} dx$

Answer 1

EXPLANATION.

$\sf \implies \displaystyle\int^{ \sqrt{ln3} } _{ \sqrt{ln2} } \dfrac{xsin(x^{2}) }{sin(x^{2} ) + sin(ln6 - x^{2} )} dx.$

As we know that,

By using substitution method, we get.

⇒ x² = t.

Differentiate w.r.t x, we get.

⇒ 2x dx = dt.

⇒ (x)dx = dt/2.

As we know that,

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

Put the lower limit

⇒ √㏑2.

⇒ (√㏑2)² = t.

⇒ ㏑2 = t. = new lower limit.

⇒ √㏑3 = t.

⇒ (√㏑3)² = t.

⇒ ㏑3 = t = upper limit.

$\sf \implies\displaystyle\int^{{ln3} } _{ {ln2} } \dfrac{sin(t)}{sint + sin(ln6 - t)}\dfrac{dt}{2}$

$\sf I \implies \dfrac{1}{2} \displaystyle\int^{{ln3} } _{ {ln2} } \dfrac{sin(t)}{sint + sin(ln6 - t)} dt \implies (1)$

As we know that

Formula of :

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

Replace x = a + b - x.

Proof :

$\sf \implies \int\limits^b_a {f(x)} \, dx$

⇒ x = a + b - t.

⇒ dx = - dt.

$\sf \implies \int\limits^a_b {f(a + b - t)} \, -dt$

a = a + b - t.

t = b.

$\sf \implies \int\limits^b_a {f(a + b - t)dt} \,$

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

HENCE PROVED.

Replace,

⇒ t = ㏑3 + ㏑2 - 1.

⇒ t = ㏑6 - 1.

$\sf I\implies\dfrac{1}{2} \displaystyle\int^{{ln3} } _{ {ln2} } \dfrac{sin(ln6 - t)}{sin(ln6 - t) + sint} dt \implies (2)$

Adding equation (1) & (2), we get.

$\sf 2I\implies\dfrac{1}{2} \displaystyle\int^{{ln3} } _{ {ln2} } \dfrac{sint + sin(ln6 - t)}{sint + sin(ln6 - t)} dt.$

$\sf 2I\implies\dfrac{1}{2} \displaystyle\int^{{ln3} } _{ {ln2} } dt$

$\sf 2I\implies\dfrac{1}{2} \bigg[ln3 - ln2\bigg]$

$\sf \implies I = \dfrac{1}{4} ln\dfrac{3}{2}$

                                                                                                                   

MORE INFORMATION.

Properties of definite integrals.

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

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

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

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

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

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\tt\implies \displaystyle\int^{ \sqrt{ln3} } _{ \sqrt{ln2} } \dfrac{xsin(x^{2}) }{sin(x^{2} ) + sin(ln6 - x^{2} )} dx$

$\tt \: Let \: I \: = \displaystyle\int^{ \sqrt{ln3} } _{ \sqrt{ln2} } \dfrac{xsin(x^{2}) }{sin(x^{2} ) + sin(ln6 - x^{2} )} dx$

$\:\boxed{\red{\bf\:Put \: {x}^{2} = t }}$

$\rm :\implies\:2xdx = dt$

$\rm :\implies\:x \: dx \: = \dfrac{dt}{2}$

Now,

• Limits become

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf t = {x}^{2} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \sqrt{ln2} & \sf ln2 \\ \\ \sf \sqrt{ln3} & \sf ln3 \end{array}} \\ \end{gathered}$

So,

• Given integral can be rewritten as

$\tt \: I = \dfrac{1}{2}\displaystyle\int^{{ln3} } _{ {ln2} } (\dfrac{sin(t)}{sint + sin(ln6 - t)})dt - - - (1)$

We know that,

$\boxed{ \pink{\int\limits^b_a {f(x)} \, dx = \int\limits^b_a{f(a + b - x)}}}$

So,

• using this property for above integral,

$\rm :\implies\:Change \: t \to \: ln3 + ln2 - t = ln6 - t$

So,

• equation (1) can be rewritten as

$\tt \: I = \dfrac{1}{2}\displaystyle\int^{{ln3} } _{ {ln2} }( \dfrac{sin(ln6 - t)}{sint + sin(ln6 - t)} )dt - - - (2)$

Now,

• Adding equation (1) and equation (2), we get

$\tt \: 2I = \dfrac{1}{2}\displaystyle\int^{{ln3} } _{ {ln2} } \dfrac{sin(t) + sin(ln - 6)}{sint + sin(ln6 - t)}dt$

$\tt \: 2I = \dfrac{1}{2}\displaystyle\int^{{ln3} } _{ {ln2} }dt$

$\rm :\implies\:\ 2I\implies\dfrac{1}{2} \bigg[ln3 - ln2\bigg]$

$\boxed{\bf \:\large\red{I=\dfrac{1}{4} ln(\dfrac{3}{2})}}$

$\large \underline{\tt \: \blue{ Explore \: more - }}$

The properties of definite integral are as follow -

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

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

$\bf (3). \: \boxed{ \blue{ \int\limits^b_a {f(x)} \, dx = - \int\limits^a_b {f(x)}}}$

$\bf (4). \: \boxed{ \purple{\int\limits^b_a {f(x)} \, dx = \int\limits^c_a {f(x)} \, dx + \int\limits^b_c {f(x)} \, dx \ where\ \ a < c < b}}$

$\bf (5). \: \boxed{ \pink{ \int\limits^a_0 {f(x)} \, dx =\int\limits^a_0 {f(a - x)}}}$

$\bf (6). \: \boxed{ \red{ \int\limits^b_a {f(x)} \, dx = \int\limits^b_a{f(a + b - x)}}}$