Let Ι_1=∫1(π)/2dt/(1+t^6) and Ι_2=∫0(π)/2(xcos⁡xdx)/(1+xsinx+cos⁡x^6), then

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) Ι_1=Ι_2. In Ι_(2,) let, xsinx+cos⁡x=t ⇒xcosx+sin⁡x-sin⁡xdx=dt or xcosxdx=dt ∴Ι_2=∫1(π)/2dt/(1+t^6)=Ι_1 Also, Ι_1 & Ι_2 are both positive as 1/(1+t^6)>0 ∀t∈1,(π)/2

Let $Ι_{1} = \int_{1}^{\frac{π}{2}} \frac{d t}{1 + t^{6}}$ and $Ι_{2} = \int_{0}^{\frac{π}{2}} \frac{x \cos x d x}{1 + {(x s i n x + \cos x)}^{6}},$ then

Solution

More Definite Integration Questions

Let Ι1=∫1π2dt1+t6 and Ι2=∫0π2xcos⁡xdx1+xsinx+cos⁡x6, then

Solution

More Definite Integration Questions

Let $Ι_{1} = \int_{1}^{\frac{π}{2}} \frac{d t}{1 + t^{6}}$ and $Ι_{2} = \int_{0}^{\frac{π}{2}} \frac{x \cos x d x}{1 + {(x s i n x + \cos x)}^{6}},$ then