Q.23 Evaluate : displaystyle∫sin(x+a)/cos(x-b) dx

Answer: cos(a+b)log|sec(x-b)|+xsin(a+b)+c Write x+a=(x-b)+(a+b) . Then sin(x+a)=sin(x-b)cos(a+b)+cos(x-b)sin(a+b) . thereforesin(x+a)/cos(x-b)=cos(a+b)tan(x-b)+sin(a+b) . I=cos(a+b)displaystyle∫tan(x-b)dx+sin(a+b)∫ dx=cos(a+b)log|sec(x-b)|+xsin(a+b)+c .

Chapter: Indefinite Integration· Mathematics & Statistics Part 2

Evaluate : \(\displaystyle\int\dfrac{\sin(x+a)}{\cos(x-b)}\,dx\)

Q.23 Evaluate : \(\displaystyle\int\dfrac{\sin(x+a)}{\cos(x-b)}\,dx\)

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

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Answer: \(\cos(a+b)\log|\sec(x-b)|+x\sin(a+b)+c\)

Write \(x+a=(x-b)+(a+b)\). Then

\(\sin(x+a)=\sin(x-b)\cos(a+b)+\cos(x-b)\sin(a+b)\).

\(\therefore\dfrac{\sin(x+a)}{\cos(x-b)}=\cos(a+b)\tan(x-b)+\sin(a+b)\).

\(I=\cos(a+b)\displaystyle\int\tan(x-b)dx+\sin(a+b)\int dx=\cos(a+b)\log|\sec(x-b)|+x\sin(a+b)+c\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Evaluate : \(\displaystyle\int\dfrac{\sin(x+a)}{\cos(x-b)}\,dx\)

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