Indefinite Integration 1 Question 55
56. Let $\frac{d}{d x} F(x)=\frac{e^{\sin x}}{x}, x>0$.
If $\int _1^{4} \frac{2 e^{\sin x^{2}}}{x} d x=F(k)-F(1)$, then one of the possible values of $k$ is …..
$(1997,2 M)$
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Solution:
- Given, $\int _1^{4} \frac{2 e^{\sin x^{2}}}{x} d x=F(k)-F(1)$
Put $\quad x^{2}=t$
$\Rightarrow \quad 2 x d x=d t$
$\Rightarrow \quad \int _1^{16} 2 \frac{e^{\sin t}}{t} \cdot \frac{d t}{2}=F(k)-F(1)$
$\Rightarrow \quad \int _1^{16} \frac{e^{\sin t}}{t} d t=F(k)-F(1)$
$$ \begin{array}{lc} \Rightarrow & {[F(t)] _1^{16}=F(k)-F(1)} \\ & \because \frac{d}{d x}{F(x)}=\frac{e^{\sin x}}{x}, \text { given } \\ \Rightarrow & F(16)-F(1)=F(k)-F(1) \\ \therefore & k=16 \end{array} $$
