Question

By using the properties of definite integrals,evaluate the integrals: $\int^\frac{\pi}{2}_{-\frac{\pi}{2}} {(x^3+x\ cos\ x+tan^5\ x+1} \, dx$ is
(A) 0
(B) 2
(C) π
(D) 1

Answer 1

Answer:

option (c) is correct

Step-by-step explanation:

Concept:

If f(x) is an odd function, then

$\int\limits^{a}_{-a}f(x)\:dx=0$

$f(x)=x^3+x\;cosx+(tanx)^5\\\\f(-x)=(-x)^3+(-x)\;cos(-x)+(tan(-x))^5\\\\f(-x)=-x^3-x\;cosx+(-tanx)^5\\\\f(-x)=-x^3-x\;cosx-tan^5x\\\\f(-x)=-[x^3+x\;cosx+tan^5x]\\\\f(-x)=-f(x)$

Therefore

f(x) is an odd function.

By property of definite integrals,

$\int\limits^{\frac{\pi}{2}}_{\frac{-\pi}{2}}[x^3+x\;cosx+tan^5x]\:dx=0$

Now,

$\int\limits^{\frac{\pi}{2}}_{\frac{-\pi}{2}}[x^3+x\;cosx+tan^5x+1]\:dx\\\\=\int\limits^{\frac{\pi}{2}}_{\frac{-\pi}{2}}[x^3+x\;cosx+tan^5x]\:dx+\int\limits^{\frac{\pi}{2}}_{\frac{-\pi}{2}}{1}\:dx\\\\=0+[x]^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\\\\=\frac{\pi}{2}-(\frac{-\pi}{2})\\\\=\frac{\pi}{2}+\frac{\pi}{2}\\\\=\pi$