Integrate with the limits √ln2 to √ln3 x*sinx^2/sinx^2+sin(ln6-x^2

1.	Integrate with the limits √ln2 to √ln3 x*sinx^2/sinx^2+sin(ln6-x^2) dx
Answer» EXPLANATION. $\sf \implies \displaystyle \int_{\sqrt{ln2} }^{\sqrt{ln3} } \dfrac{x sin(x^{2} )}{sin(x^{2} ) + sin(ln 6 - x^{2} )} <klux>DX</klux>$ As we know that, We can use substitution method in this equation, we get. Let, we assume that. ⇒ 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. First, we put the lower limit, we get. ⇒ (√㏑2)² = t. ⇒ ㏑2 = t. = [NEW lower limit]. ⇒ (√㏑3)² = t. ⇒ ㏑3 = t. = [new UPPER limit]. Put the values in the equation, we get. $\sf \implies \displaystyle \int_{{ln2} }^{{ln3} } \dfrac{sin(t)}{sin(t) + sin(ln 6 - t)} \dfrac{dt}{2}$ $\sf \implies I = \dfrac{1}{2} \displaystyle \int_{{ln2} }^{{ln3} } \dfrac{sin(t)}{sin(t) + sin(ln 6 - t)} dt. - - - - - (1).$ As we know that, Formula of : $\sf \implies \int\limits^b_a {f(x)} \, dx = \int\limits^b_a {f(a + b - x)} \, dx$ Proof : REPLACE : x = a + b - x. $\sf \implies \int\limits^b_a {f(x)} \, dx$ ⇒ x = a + b - x. ⇒ dx = - dt. $\sf \implies \int\limits^a_b {f(a + b - x)} \,- dt$ a = a + b - t. $\sf \implies \int\limits^b_a {f(a+ b - t) } \, dt$ Replace t = x in the equation, we get. $\sf \implies \int\limits^b_a {f(a + b - x)} \, dx$ Hence Proved. Replace, ⇒ t = ㏑3 + ㏑2 - t. ⇒ t = ㏑6 - t. $\sf \implies I = \dfrac{1}{2} \displaystyle \int_{{ln2} }^{{ln3} } \dfrac{sin(ln 6 - t)}{sin(ln 6 - t) + sin(t)} dt. - - - - - (2).$ Adding equation (1) and (2), we get. $\sf \implies 2I = \dfrac{1}{2} \displaystyle \int_{{ln2} }^{{ln3} } \dfrac{sin(t) + sin(ln 6 - t)}{sin(t )+ sin(ln 6 - t)} dt.$ $\sf \implies 2I = \dfrac{1}{2} \displaystyle \int_{{ln2} }^{{ln3} } dt.$ Putt the upper and lower limit in the equation, we get. $\sf \implies 2I = \dfrac{1}{2} \displaystyle \bigg[t \bigg]_{ln 2}^{ln 3}$ $\sf \implies 2I = \dfrac{1}{2} \displaystyle \bigg[ ln 3 - ln 2 \bigg]$ $\sf \implies I = \dfrac{1}{4} ln \dfrac{3}{2}$ MORE INFORMATION. Properties of definite integration. $\sf (1) = \int\limits^b_a {f(x)} \, dx = \int\limits^b_a {f(t)} \, dt$ $\sf (2) = \int\limits^b_a {f(x)} \, dx = - \int\limits^a_b {f(x)} \, dx$ $\sf (3) = \int\limits^b_a {f(x)} \, dx = \int\limits^c_a {f(x)} \, dx + \int\limits^b_c {f(x)} \, dx \ \ where \ \ a < c < b.$ $\sf (4) = \int\limits^a_0 {f(x)} \, dx = \int\limits^a_0 {f(a - x)} \, dx$ $\sf (5) = \int\limits^b_a {f(x)} \, dx = \int\limits^b_a {f(a + b - x)} \, dx$

Integrate with the limits √ln2 to √ln3 x*sinx^2/sinx^2+sin(ln6-x^2) dx

Answer»

EXPLANATION.

$\sf \implies \displaystyle \int_{\sqrt{ln2} }^{\sqrt{ln3} } \dfrac{x sin(x^{2} )}{sin(x^{2} ) + sin(ln 6 - x^{2} )} <klux>DX</klux>$

As we know that,

We can use substitution method in this equation, we get.

Let, we assume that.

⇒ 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.

First, we put the lower limit, we get.

⇒ (√㏑2)² = t.

⇒ ㏑2 = t. = [NEW lower limit].

⇒ (√㏑3)² = t.

⇒ ㏑3 = t. = [new UPPER limit].

Put the values in the equation, we get.

$\sf \implies \displaystyle \int_{{ln2} }^{{ln3} } \dfrac{sin(t)}{sin(t) + sin(ln 6 - t)} \dfrac{dt}{2}$

$\sf \implies I = \dfrac{1}{2} \displaystyle \int_{{ln2} }^{{ln3} } \dfrac{sin(t)}{sin(t) + sin(ln 6 - t)} dt. - - - - - (1).$

As we know that,

Formula of :

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

Proof :

REPLACE : x = a + b - x.

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

⇒ x = a + b - x.

⇒ dx = - dt.

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

a = a + b - t.

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

Replace t = x in the equation, we get.

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

Hence Proved.

Replace,

⇒ t = ㏑3 + ㏑2 - t.

⇒ t = ㏑6 - t.

$\sf \implies I = \dfrac{1}{2} \displaystyle \int_{{ln2} }^{{ln3} } \dfrac{sin(ln 6 - t)}{sin(ln 6 - t) + sin(t)} dt. - - - - - (2).$

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

$\sf \implies 2I = \dfrac{1}{2} \displaystyle \int_{{ln2} }^{{ln3} } \dfrac{sin(t) + sin(ln 6 - t)}{sin(t )+ sin(ln 6 - t)} dt.$

$\sf \implies 2I = \dfrac{1}{2} \displaystyle \int_{{ln2} }^{{ln3} } dt.$

Putt the upper and lower limit in the equation, we get.

$\sf \implies 2I = \dfrac{1}{2} \displaystyle \bigg[t \bigg]_{ln 2}^{ln 3}$

$\sf \implies 2I = \dfrac{1}{2} \displaystyle \bigg[ ln 3 - ln 2 \bigg]$

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

MORE INFORMATION.

Properties of definite integration.

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

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

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

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

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