Question

$\int \tan\frac{\theta}{2}\sin\theta\cos\theta\, d\theta$ is equal to

• A. $\sin\theta-\frac{1}{2}\theta-\frac{\sin 2\theta}{6}+C$
• B. $\sin\theta-\frac{1}{2}\theta-\frac{\sin 2\theta}{2}+C$
• C. $2\sin\theta-\frac{1}{2}\theta-\frac{\sin 2\theta}{4}+C$
• D. $\sin\theta-\frac{1}{2}\theta-\frac{\sin 2\theta}{4}+C$
• E. $\sin\theta-\frac{1}{4}\theta-\frac{\sin 2\theta}{4}+C$

Hint:

Math Tip: When an integral mixes half-angles ($\frac{\theta}{2}$) and full angles ($\theta$), always use double-angle formulas ($\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$) to collapse the terms into a uniform angle before integrating.

Answer 1

The Correct Option is D

Concept:
Calculus - Integration using Trigonometric Identities.
Step 1: Simplify the trigonometric expression.
Rewrite $\tan\left(\frac{\theta}{2}\right)$ in terms of sine and cosine: $$ \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta/2)}{\cos(\theta/2)} $$ Apply the double-angle identity to $\sin\theta$: $$ \sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) $$
Step 2: Substitute and cancel terms.
Multiply the rewritten terms: $$ \tan\left(\frac{\theta}{2}\right) \sin\theta = \frac{\sin(\theta/2)}{\cos(\theta/2)} \cdot 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) $$ $$ = 2\sin^2\left(\frac{\theta}{2}\right) $$ Use the half-angle identity $2\sin^2\left(\frac{\theta}{2}\right) = 1 - \cos\theta$: $$ \tan\left(\frac{\theta}{2}\right) \sin\theta = 1 - \cos\theta $$
Step 3: Construct the simplified integral.
Substitute this simplified expression back into the original integral: $$ \int (1 - \cos\theta) \cos\theta \,d\theta $$ $$ = \int (\cos\theta - \cos^2\theta) \,d\theta $$
Step 4: Apply the power-reduction formula.
Rewrite $\cos^2\theta$ using the identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$: $$ \int \left( \cos\theta - \frac{1}{2} - \frac{\cos(2\theta)}{2} \right) d\theta $$
Step 5: Integrate term by term.
$$ = \sin\theta - \frac{1}{2}\theta - \frac{\sin(2\theta)}{2 \times 2} + C $$ $$ = \sin\theta - \frac{1}{2}\theta - \frac{\sin(2\theta)}{4} + C $$