Analyzing the Setup
The integral provided is:
I=∫0π(1+cos2x)(ecosx+e−cosx)ecosxsinxdx
This expression appears intimidating due to the combination of exponential and trigonometric functions. However, in JEE Advanced mathematics, such complexity is often a signal to look for symmetry.
The King's Property
Whenever you encounter a definite integral with limits from 0 to a, the 'King's Property' is your most powerful tool:
Applying this to our integral by replacing x with π−x, we obtain:
I=∫0π(1+cos2(π−x))(ecos(π−x)+e−cos(π−x))ecos(π−x)sin(π−x)dx
The Algebraic Dance
Using the trigonometric identities sin(π−x)=sinx and cos(π−x)=−cosx, the expression simplifies to:
I=∫0π(1+cos2x)(e−cosx+ecosx)e−cosxsinxdx
Notice that the denominator remains unchanged. By adding the original integral I to this new form, the exponential terms in the numerator and denominator cancel out:
2I=∫0π(1+cos2x)(ecosx+e−cosx)sinx(ecosx+e−cosx)dx
This reduces the problem to a much simpler form:
The Final Descent
Since the integrand is symmetric about π/2, we can simplify the integral to:
We now perform the substitution t=cosx, which implies dt=−sinxdx. Adjusting the limits, when x=0, t=1, and when x=π/2, t=0:
I=∫101+t2−dt=∫011+t2dt
Evaluating this standard integral yields:
I=[tan−1(t)]01=tan−1(1)−tan−1(0)
The final result is: