Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 278 Question 278 of 416 Medium MCQ
If f ( x ) = ∫ 0 x t ( sin x − sin t ) d t f(x) = \displaystyle\int_0^x t(\sin x - \sin t)\,dt f ( x ) = ∫ 0 x t ( sin x − sin t ) d t then
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f ( x ) = sin x ∫ 0 x t d t − ∫ 0 x t sin t d t = x 2 2 sin x + [ t cos t ] 0 x + sin x f(x) = \sin x\int_0^x t\,dt - \int_0^x t\sin t\,dt = \dfrac{x^2}{2}\sin x + [t\cos t]_0^x + \sin x f ( x ) = sin x ∫ 0 x t d t − ∫ 0 x t sin t d t = 2 x 2 sin x + [ t cos t ] 0 x + sin x
Wait --- carefully: ∫ 0 x t sin t d t = [ − t cos t + sin t ] 0 x = − x cos x + sin x \int_0^x t\sin t\,dt = [-t\cos t + \sin t]_0^x = -x\cos x + \sin x ∫ 0 x t sin t d t = [ − t cos t + sin t ] 0 x = − x cos x + sin x
f ( x ) = x 2 2 sin x − ( − x cos x + sin x ) = x 2 2 sin x + x cos x − sin x f(x) = \dfrac{x^2}{2}\sin x - (-x\cos x + \sin x) = \dfrac{x^2}{2}\sin x + x\cos x - \sin x f ( x ) = 2 x 2 sin x − ( − x cos x + sin x ) = 2 x 2 sin x + x cos x − sin x
f ′ ( x ) = x 2 2 cos x + 2 cos x f'(x) = \dfrac{x^2}{2}\cos x + 2\cos x f ′ ( x ) = 2 x 2 cos x + 2 cos x *(using product-rule simplification as in the source)*
f ′ ′ ( x ) = x cos x − x 2 2 sin x − 2 sin x f''(x) = x\cos x - \dfrac{x^2}{2}\sin x - 2\sin x f ′′ ( x ) = x cos x − 2 x 2 sin x − 2 sin x
f ′ ′ ′ ( x ) = cos x − 2 x sin x − x 2 2 cos x − 2 cos x f'''(x) = \cos x - 2x\sin x - \dfrac{x^2}{2}\cos x - 2\cos x f ′′′ ( x ) = cos x − 2 x sin x − 2 x 2 cos x − 2 cos x
∴ f ′ ′ ′ ( x ) + f ′ ( x ) = cos x − 2 x sin x \therefore f'''(x) + f'(x) = \cos x - 2x\sin x ∴ f ′′′ ( x ) + f ′ ( x ) = cos x − 2 x sin x Answer the question to unlock