Integration

Calculus Level 2

∫ x 2 sin ⁡ x d x = ? \large \int x^2 \sin x \ dx = \, ?

Clarification : C C denotes the arbitrary constant of integration .

3 cos ⁡ x sin ⁡ x + sin ⁡ 2 x + C 3 \cos x \sin x + \sin^2 x + C 2 x sin ⁡ x + 2 cos ⁡ x − x 2 cos ⁡ x + C 2x\sin x + 2\cos x - x^2\cos x + C 2 x sin ⁡ x + cos ⁡ x + C 2x \sin x + \cos x + C sin ⁡ 2 x − 9 cos ⁡ x + tan ⁡ x + C \sin 2x - 9 \cos x + \tan x + C

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Abdullah Mughal by Abdullah Mughal , 18, Switzerland

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1 solution

Chew-Seong Cheong
Jun 14, 2016

By integration by parts:

I = ∫ x 2 sin ⁡ x d x f ′ = sin ⁡ x , g = x 2 = − x 2 cos ⁡ x + ∫ 2 x cos ⁡ x d x = − x 2 cos ⁡ x + 2 x sin ⁡ x − ∫ 2 sin ⁡ x d x = − x 2 cos ⁡ x + 2 x sin ⁡ x + 2 cos ⁡ x + c o n s t a n t = 2 x sin ⁡ x + 2 cos ⁡ x − x 2 cos ⁡ x + c o n s t a n t \begin{aligned} I & = \int x^2 \sin x \ dx \quad \quad \small \color{#3D99F6}{f' = \sin x, \ g = x^2} \\ & = - x^2 \cos x + \int 2x \cos x \ dx \\ & = - x^2 \cos x + 2x \sin x - \int 2\sin x \ dx \\ & = - x^2 \cos x + 2x \sin x + 2\cos x + \small \color{grey}{constant} \\ & = \boxed {2x \sin x + 2\cos x - x^2 \cos x + \small \color{grey}{constant}} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Abdullah Mughal , you have provided a wrong answer earlier. I have edited the problem and the answer options for you.

Chew-Seong Cheong - 5 years ago

Thank you for spotting the mistake.

Abdullah Mughal - 5 years ago

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