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Indefinite Integration

Section: Mathematics

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Question : 21 of 89

Marks: +1, -0

If

\int e^{\sin x}(1+\sec x\tan x) dx=e^{\sin x} f(x)+c

0 \le x \le 2 \pi

, the number of solutions of

f(x)=1

[AP EAPCET 21 May 2025 S2]

4
0
2
3

Solution:

\, \text{ Given }\, \int e^{\sin x}(1+\sec x\tan x) dx

= e^{\sin x} f(x)+c

\int e^{\sin x} \left(\, \frac{1}{\cos x}+\, \frac{\sin x}{\cos^3 x}\,\right) \cos x dx

Put

\sin x = t

\, \cos x \, dx = dt

\, \cos x = \sqrt{1-t^2}

\, \int e^t \left[ \frac{1}{\sqrt{1-t^2}} + \frac{t}{(1-t^2)^{3/2}} \right] \, dt

[Using result

\int e^x \left[ f(x) + f'(x) \right] \, dx

= e^x f(x)+C]

\int e^t \left[ \frac{1}{\sqrt{1-t^2}} + \left( \frac{1}{\sqrt{1-t^2}} \right)^1 \right] \, dt = e^t \cdot \frac{1}{\sqrt{1-t^2}}

= e^{\sin x} \cdot \frac{1}{\cos x} = e^{\sin x} \cdot \sec x + C

Comparing with

e^{\sin x} f(x)+C

\therefore \, \, f(x) \, = \, \sec x = 1

x \, = 0, 2 \pi

\therefore

Number of solution is 2 .

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