The integral I = ∫ [xe^ x2 (sin x^ 2 + cos x^ 2 )]dx= f (x) + c, (where, c is the constant of integration). Then, f(x) can be

Correct answer: D. (1)/(2)e^ x^ 2 (x^ 2 ) Let,& x^ 2 = t ⇒ 2xdx = dt ∴ I = (1)/(2)∫ e^ t (sin t + cos t)dt = (1)/(2)e^ t ⋅ sin t + c = (1)/(2)e^ x^ 2 (x^ 2 ) + c \ As, ∫ e^ x (f(x) + f^ ′ (x))dx = e^ x ⋅ f(x) + c\

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The integral I = ∫ [xe^ x2 (sin x^ 2 + cos x^ 2 )]dx= f (x) + c,...

The integral

I = ∫ \left[xe^{x2}\left(sin x^{2} + cos x^{2}\right)\right]dx= f \left(x\right) + c,

(where,

c

is the constant of integration). Then,

f\left(x\right) can be

Options

A $e^{x}\sin \left(x^{2}\right)$
B $e^{x^{2}}\sin \left(x\right)$
C $e^{x^{2}}\sin \left(\dfrac{x^{2}}{2}\right)$
D $\dfrac{1}{2}e^{x^{2}}\sin \left(x^{2}\right)$

Correct answer

D. $\dfrac{1}{2}e^{x^{2}}\sin \left(x^{2}\right)$

Step-by-step solution

Let,&

x^{2} = t ⇒ 2xdx = dt

∴ I = \dfrac{1}{2}∫ e^{t}\left(sin t + cos t\right)dt

= \dfrac{1}{2}e^{t} ⋅ sin t + c

= \dfrac{1}{2}e^{x^{2}}\sin \left(x^{2}\right) + c

\left\{\text{As,} ∫ e^{x} \left(f\left(x\right) + f^{′} \left(x\right)\right)dx = e^{x} ⋅ f\left(x\right) + c\right\}

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