Question

integral e^ (log x + log sin x )dx​

Answer 1

Question :

Integrate

$\tt\int\:e^{(\log\:x+\log\:sin\:x)}dx$

Solution :

We have to integrate

$\sf\int\:e^{(\log\:x+\log\:sin\:x)}dx$

We know that:

$\sf\:\log\:a+\log\:b=\log(ab)$

$\sf\:and\:e^{\log_e(a)}=a$

Thus ,

$\sf\int\:e^{(\log\:x+\log\:sin\:x)}dx$

$\sf\int\:e^{\log(x\:\sin\:x)}dx$

$\sf\int\:x\:\sin\:x\:dx$

Now , Apply Integration by parts :

Then ,

$\sf\int\:x\:\sin\:x\:dx$

$\sf\:=x\times(-\cos\:x)-\int\:(-cosx)\:x\:dx$

$\sf\:=x\times(-\cos\:x)-(-sinx)+c$

$\sf\:=-x\cos\:x+\sin\:x+c$

Hence ,

$\tt\int\:e^{(\log\:x+\log\:sin\:x)}dx=-x\cos\:x+\sin\:x+c$

More About the topic :

Integral by part:

$\tt\int\:u(x)\times\:v(x)=u(x)\int\:v(x)dx-\int[u'(x)\times\int\:v(x)dx]$

Where , u(x) and v(x) are diffentiable function .

Answer 2

Given :

• $∫ e^{log x + log\:sin x}dx$

To find :

• $∫ e^{log x + log\:sin x}dx$

According to the question :

$∫ e^{log x + log\:sin x}dx$

$∫ x sin ( x ) dx$

Using partial Integration Formula :

⟹ $∫ udv = uv - ∫ vdu$

⟹ $u ( x ) = x, dv = ( x )$

Solving :

⟹ $-x cos ( x ) - ∫ ( -cos ( x )) dx$

⟹ $∫ af ( x ) dx = a ∫ f ( x ) dx$

⟹ $-x cos ( x ) - ( - ∫ cos ( x ) dx )$

→ The integral of cosine is sine.

⟹ $-x cos ( x ) - ( -sin ( x ))$

⟹ $-x cos ( x ) + sin ( x )$

⟹ $-x cos ( x ) + sin ( x ) + c$

How did I solved this :

⇛Used partial Integration Formula

⇛Used Property of Integration.

So, It's Done !!