Animated Solution for Mathematics - Indefinite Integration: The value of the integral ∫2sin4θ+3sin2θ+6sinθ−sin2θ(sin3θ+sin2θ+sinθ)1−cos2θdθ is (where c is a constant of integration)
Select Answer:
Visualized Solution
The Reverse Engineering Strategy
The given integral looks extremely complex and algebraically heavy.
JEE Trick: Look at the options before starting!
All options have the form 181[f(θ)]23+c.
This suggests the integral eventually simplifies to the form ∫zdz.
Extracting z from Options
Let's extract the core polynomial from the options.
Let z=2sin6θ+3sin4θ+6sin2θ.
Differentiating the Substitution
Differentiate z with respect to θ:
dz=(12sin5θcosθ+12sin3θcosθ+12sinθcosθ)dθ
Simplifying dz
Factor out 12cosθ:
dz=12cosθ(sin5θ+sin3θ+sinθ)dθ
This perfectly matches the simplified numerator of the original integral!
Transforming the Integral
The entire complex integral reduces to:
I=121∫zdz
Power Rule Integration
Apply power rule: ∫z21dz=23z23
I=121⋅32z23+c
I=181z23+c
Substituting z Back
Substitute z=2sin6θ+3sin4θ+6sin2θ:
I=181[2sin6θ+3sin4θ+6sin2θ]23+c
Sine to Cosine Conversion
Options 3 and 4 use cosθ.
Substitute sin2θ=1−cos2θ into z.
z=2(1−cos2θ)3+3(1−cos2θ)2+6(1−cos2θ)
Algebraic Expansion
Expand using (1−x)3=1−3x+3x2−x3
2(1−3cos2θ+3cos4θ−cos6θ)
+3(1−2cos2θ+cos4θ)
+6−6cos2θ
Combining Like Terms
Constants: 2(1)+3(1)+6=11
cos2θ: 2(−3)+3(−2)−6=−18
cos4θ: 2(3)+3(1)=9
cos6θ: 2(−1)=−2
z=11−18cos2θ+9cos4θ−2cos6θ
Final Answer
Final Integral: 181[11−18cos2θ+9cos4θ−2cos6θ]23+c
At first glance, this expression appears daunting due to the trigonometric complexity. However, in JEE Advanced, the provided options often serve as a roadmap. Since every option is in the form 181[f(θ)]23+c, we can infer that the integral is a disguised version of the power rule:
∫z21dz
The Detective Work
We extract the core polynomial from the structure of the options. Let us define:
z=2sin6θ+3sin4θ+6sin2θ
If our hypothesis is correct, the derivative of z must be hidden within the numerator of the integral. We differentiate z with respect to θ using the chain rule:
dz=(12sin5θcosθ+12sin3θcosθ+12sinθcosθ)dθ
Factoring out 12cosθ, we obtain:
dz=12cosθ(sin5θ+sin3θ+sinθ)dθ
This confirms that the complexity of the original integral was merely a mask. By identifying z from the options, we bypass tedious simplification and reduce the integral to:
I=121∫zdz
The Final Transformation
Applying the power rule ∫z21dz=23z23, and multiplying by our constant 121, we get:
I=121⋅32z23+c=181z23+c
Substituting our original definition of z back into the expression, we have:
I=181[2sin6θ+3sin4θ+6sin2θ]23+c
To match the specific form of the provided options, we use the identity sin2θ=1−cos2θ. Substituting this into z and expanding the terms, we arrive at:
z=11−18cos2θ+9cos4θ−2cos6θ
This result matches the required form perfectly. By using the options as a strategic map, we have successfully solved the problem. The final answer is derived from the substitution of this polynomial into the power rule result.