Indefinite Integration
Rational trig integral — constants A and B
MJAT_TS8_P2
Grade 12
Question:
If $\displaystyle\int\frac{\sin x}{(x+4)(\sin x-1)}\,dx=A\tan^{-1}\!\left(\frac{4\tan^2x+15}{1}\right)+B\tan^{-1}(\cdots)+C$, then:
A) $A=\dfrac{2}{5}$
B) $B=-\dfrac{2}{5\sqrt{15}}$
C) $A=-\dfrac{2}{5}$
D) $B=1$
Step-by-Step Solution
Key Concept: Use the substitution for $\int\frac{\sin x}{(\sin x+4)(\sin x-1)}dx$. Partial fractions: $\frac{1}{5}\int\frac{1}{\sin x-1}dx-\frac{1}{5}\int\frac{1}{\sin x+4}dx$. Each integral is computed by Weierstrass substitution $t=\tan(x/2)$.
A ✓, B ✓. Answer: A, B.
Correct Answer: AB