Integral Calculus-2
Integral Calculus-2
Allen Star Batch
Grade 12
Question:
If the absolute value of the integral $I = \int_{\pi/4}^{\pi/2} \frac{x \cdot \cos 2x \cdot \cos x}{\sin^3 x} dx$ in the lowest form is $-\frac{a}{b}$ where $a,b \in \mathbb{N}$, then $(a+b) = $ ____.
Step-by-Step Solution
Key Concept: Decompose the numerator using double angle formulas, then apply substitution on the denominator to obtain inverse powers of sine.
For $I_1 = \int \frac{\cos 2x \cos x}{\sin^3 x} dx$, use the identity $(1-2\sin^2 x)\cos x = \cos x - 2\sin^2 x \cos x$. With substitution $t = \sin x$, we get $I_1 = \int \frac{1-2t^2}{t^3} dt = -\frac{1}{2t^2} + 2\ln|t| = -\frac{1}{2\sin^2 x} + 2\ln|\sin x| + C$. Express as $\frac{\cosec^4 x}{2} - \frac{\cosec^6 x}{6}$ using trigonometric identities.
Correct Answer: 61