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REPEATER 2024 — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsREPEATER 20243 Marks
Question
Evaluate: $\int \frac{\sin (x+a)}{\cos (x-b)} d x$
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Answer

Let $I=\int \frac{\sin (x+a)}{\cos (x-b)} d x$
$\therefore I=\int \frac{\sin [(x-b)+(a+b)]}{\cos (x-b)} d x$
$=\int \frac{\sin (x-b) \cdot \cos (a+b)+\cos (x-b) \cdot \sin (a+b)}{\cos (x-b)} d x \quad \ldots . .[\because \quad \sin (A+B)=\sin A \cdot \cos B+\cos A \cdot \sin B]$
$=\int\left[\frac{\sin (x-b) \cdot \cos (a+b)}{\cos (x-b)}+\frac{\cos (x-b) \cdot \sin (a+b)}{\cos (x-b)}\right] d x$
$=\int[\tan (x-b) \cdot \cos (a+b)+\sin (a+b)] d x$
$=\cos (a+b) \int \tan (x-b) \cdot d x+\sin (a+b) \int d x$
$I=\cos (a+b) \cdot \log |\sec (x-b)|+x \cdot \sin (a+b)+c \quad \ldots \ldots \quad\left[\because \quad \int \tan x \cdot d x=\log |\sec x|+c\right]$

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