Question

integrate 3x ^ 2 * sin(x ^ 3 + 1) dx​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\tt {3x}^{2} sin \: ( {x}^{3} + 1) \: dx$

Now, we use here method of Substitution,

$\red{\rm :\longmapsto\:Put \: {x}^{3} + 1 = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto\:\dfrac{d}{dx}( {x}^{3} + 1) = \dfrac{d}{dx}y}$

$\red{\rm :\longmapsto\: {3x}^{2} = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: {3x}^{2}dx = dy}$

On substituting all these values in given integral, we get

$\rm \: = \: \: \displaystyle\int\tt siny \: dy$

$\rm \: = \: \: - \: cosy \: + \: c$

$\tt \: = \: \: - \: cos( {x}^{3} + 1)\: + \: c$

Additional Information :-

Let's solve few more examples of same type!!!

Question: - 1

$\red{\rm :\longmapsto\:\displaystyle\int\tt {e}^{tanx} {sec}^{2} x \: dx}$

To solve this integral, we use method of Substitution

$\red{\rm :\longmapsto\:Put \: tanx = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto \: \dfrac{d}{dx} \: tanx =\dfrac{d}{dx} y}$

$\red{\rm :\longmapsto\: {sec}^{2}x \: dx \: = \: dy}$

So, above integral can be rewritten as

$\rm \: = \: \: \displaystyle\int\tt {e}^{y} \: dy$

$\tt \: = \: \: {e}^{y} \: + \: c$

$\tt \: = \: \: {e}^{tanx} \: + \: c$

Question :- 2

$\red{\rm :\longmapsto\:\displaystyle\int\tt \dfrac{(x + 1)cos(x + logx)}{x} \: dx}$

To solve this integral, we use method of Substitution

$\red{\rm :\longmapsto\:Put \: x + logx = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto\:\dfrac{d}{dx} \: (x + logx) = \dfrac{d}{dx}y}$

$\red{\rm :\longmapsto\:1 + \dfrac{1}{x} = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: \dfrac{x + 1}{x} = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: \dfrac{x + 1}{x} \: dx =dy}$

So, above integral can be rewritten as

$\rm \: = \: \: \displaystyle\int\tt cosy \: dy \:$

$\tt \: = \: \: siny \: + \: c$

$\tt \: = \: \: sin(x + logx) \: + \: c$