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Exercise 9.10 · Q3

Q.For any value of n∈Zn\in\mathbb{Z}, ∫0πecos⁡2xcos⁡3[(2n+1)x] dx\displaystyle\int_0^\pi e^{\cos^2x}\cos^3\big[(2n+1)x\big]\,dx is

(1) π2\dfrac{\pi}{2}
(2) π\pi
(3) 00
(4) 22
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✓ Free question

Applying Property 6 (∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x)dx=\int_a^b f(a+b-x)dx) with a=0,b=πa=0,b=\pi shows the integrand reverses sign under x→π−xx\to\pi-x, which forces the integral to equal its own negative — hence 00.

Step 1. Name the integral. Let I=∫0πecos⁡2xcos⁡3[(2n+1)x] dxI=\displaystyle\int_0^\pi e^{\cos^2x}\cos^3[(2n+1)x]\,dx.

Step 2. Replace xx by π−x\pi-x (Property 6, a+b−xa+b-x trick with a=0,b=πa=0,b=\pi).

cos⁡2(π−x)=(−cos⁡x)2=cos⁡2x\cos^2(\pi-x)=(-\cos x)^2=\cos^2x, so ecos⁡2(π−x)=ecos⁡2xe^{\cos^2(\pi-x)}=e^{\cos^2x} is unchanged.

Step 3. Simplify cos⁡[(2n+1)(π−x)]\cos[(2n+1)(\pi-x)]. Expand: cos⁡[(2n+1)π−(2n+1)x]=cos⁡[(2n+1)π]cos⁡[(2n+1)x]+sin⁡[(2n+1)π]sin⁡[(2n+1)x]\cos[(2n+1)\pi-(2n+1)x]=\cos[(2n+1)\pi]\cos[(2n+1)x]+\sin[(2n+1)\pi]\sin[(2n+1)x].

Since 2n+12n+1 is an odd integer, cos⁡[(2n+1)π]=−1\cos[(2n+1)\pi]=-1 and sin⁡[(2n+1)π]=0\sin[(2n+1)\pi]=0, so

cos⁡[(2n+1)(π−x)]=−cos⁡[(2n+1)x].\cos[(2n+1)(\pi-x)]=-\cos[(2n+1)x].

Cubing: cos⁡3[(2n+1)(π−x)]=−cos⁡3[(2n+1)x]\cos^3[(2n+1)(\pi-x)]=-\cos^3[(2n+1)x].

Step 4. Combine. So the integrand at π−x\pi-x is ecos⁡2x⋅(−cos⁡3[(2n+1)x])e^{\cos^2x}\cdot\left(-\cos^3[(2n+1)x]\right), i.e. exactly the negative of the original integrand.

Step 5. Apply Property 6 and solve for II.

I=∫0πecos⁡2xcos⁡3[(2n+1)x] dx=∫0π(−ecos⁡2xcos⁡3[(2n+1)x])dx=−I.I=\int_0^\pi e^{\cos^2x}\cos^3[(2n+1)x]\,dx=\int_0^\pi \left(-e^{\cos^2x}\cos^3[(2n+1)x]\right)dx=-I.

So 2I=0⇒I=02I=0\Rightarrow I=0, for every integer nn.

Step 6. Match to the printed options. 00 is option (3).

✓Final answer

Option (3): 00.

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