Choose the correct answer- If f x - x 0 t sin t dt- then f - x isa cos x-x sin xb x sin xc x cos xdsin x-x cos x

Given, f(x)=∫_0^xt sin t dt On integrating by parts, we get f(x)=t∫_0^x sin t dt ∫_0^x [ (d/dtt)∫ sin t dt ]dt =[t( cos t)]^x_0 ∫_0^x ( cos t)dt =[ t cos t ...

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Byju's Answer

Standard XII

Mathematics

Integration Using Substitution

Choose the co...

Question

Choose the correct answer.
If $f (x) = \int_{x}^{0}$ t sin t dt, then f'(x)is
(a)cos x+x sin x
(b)x sin x
(c)x cos x
(d)sin x +x cos x

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Solution

Given, $f (x) = \int_{0}^{x} t s i n t d t$
On integrating by parts, we get
$f (x) = t \int_{0}^{x} s i n t d t - \int_{0}^{x} [(\frac{d}{d t} t) \int s i n t d t] d t = [t (- c o s t)]_{0}^{x} - \int_{0}^{x} (- c o s t) d t = [- t c o s t + s i n t]_{0}^{x} \Rightarrow f (x) = - x c o s x + s i n x$
Now, differentiate w.r.t x,we get
$f'(x)=-[{x(-sin x)}+cos x]+cos x=x sin x-cos x +cos x =x sin x$
Using the product rule $\frac{d}{d x} (u . v) = u \frac{d v}{d x} + v \frac{d u}{d x}$
Hence, option (b) is the correct answer.

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Choose the correct answer. If f(x)=∫0x t sin t dt, then f'(x)is (a)cos x+x sin x (b)x sin x (c)x cos x (d)sin x +x cos x

Choose the correct answer.
If $f (x) = \int_{x}^{0}$ t sin t dt, then f'(x)is
(a)cos x+x sin x
(b)x sin x
(c)x cos x
(d)sin x +x cos x