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Maths Question 18 – JEE-MAIN 2026

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The value of the integral ∫−π4π4(32cos4⁡x1+esin⁡x)dx is:

For an integral of the form ∫−aaf(x)dx, consider using the property ∫−aaf(x)dx=∫0a[f(x)+f(−x)]dx.

● Video Walkthrough
Step 1: Apply Definite Integral Property✦ Active

Let the given integral be I. We use the property ∫−aaf(x)dx=∫0a[f(x)+f(−x)]dx. Here, a=π4 and f(x)=32cos4⁡x1+esin⁡x. We find f(−x):

f(−x)=32cos4⁡(−x)1+esin⁡(−x)=32cos4⁡x1+e−sin⁡x=32cos4⁡x1+1esin⁡x=32cos4⁡x⋅esin⁡xesin⁡x+1

Now, sum f(x) and f(−x):

f(x)+f(−x)=32cos4⁡x1+esin⁡x+32cos4⁡x⋅esin⁡x1+esin⁡x=32cos4⁡x(1+esin⁡x)1+esin⁡x=32cos4⁡x

So the integral simplifies to:

I=∫0π432cos4⁡xdx
Step 2: Simplify the Integrand using Trigonometric Identities○ Expand

We use the power reduction formula cos2⁡θ=1+cos⁡(2θ)2 to simplify cos4⁡x:

cos4⁡x=(cos2⁡x)2=(1+cos⁡(2x)2)2=1+2cos⁡(2x)+cos2⁡(2x)4

Apply the formula again for cos2⁡(2x):

cos2⁡(2x)=1+cos⁡(4x)2

Substitute back into the expression for cos4⁡x:

cos4⁡x=1+2cos⁡(2x)+1+cos⁡(4x)24=2+4cos⁡(2x)+1+cos⁡(4x)8=3+4cos⁡(2x)+cos⁡(4x)8
💡 Teacher's Secret Hint

Remember to apply power reduction formulas carefully to avoid errors in coefficients.

Step 3: Evaluate the Definite Integral○ Expand

Substitute the simplified cos4⁡x back into the integral:

I=32∫0π43+4cos⁡(2x)+cos⁡(4x)8dx=4∫0π4(3+4cos⁡(2x)+cos⁡(4x))dx

Integrate term by term:

I=4[3x+4sin⁡(2x)2+sin⁡(4x)4]0π4=4[3x+2sin⁡(2x)+14sin⁡(4x)]0π4

Evaluate at the limits:

I=4[(3(π4)+2sin⁡(2⋅π4)+14sin⁡(4⋅π4))−(3(0)+2sin⁡(0)+14sin⁡(0))]
I=4[(3π4+2sin⁡(π2)+14sin⁡(π))−(0)]
I=4[3π4+2(1)+14(0)]=4[3π4+2]=3π+8

The value of the integral is 3π+8.

💡 Teacher's Secret Hint

Be careful with the evaluation of trigonometric functions at the limits, especially sin⁡(π/2) and sin⁡(π).

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