Question

Assertion :${\int }_0^{\pi /4}\frac{\cos x+\sin x}{{\cos }^{2}x+{\sin }^{4}x}dx=\frac{\pi }{4}+\frac{1}{2\sqrt{3}}\log (2+\sqrt{3})=I$ Reason: $I={\int }_0^1\frac{dx}{1-x^2+x^4}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$I={\int }_0^{\pi /4}\frac{\cos x+\sin x}{{\cos }^{2}x+{\sin }^{4}x}dx$
$I={\int }_0^{\pi /4}\frac{\cos x}{{\cos }^{2}x+{\sin }^{4}x}+{\int }_0^{\pi /4}\frac{\sin x}{{\cos }^{2}x+{\sin }^{4}x}$
$I={\int }_0^{\pi /4}\frac{\cos xdx}{1-{\sin }^{2}x+{\sin }^{4}x}+{\int }_0^{\pi /4}\frac{\sin xdx}{{\cos }^{4}x-{\cos }^{2}x+1}$
Put $\sin x=t$ in the first integral $\Rightarrow \cos xdx=dt$
And put $\cos x=u$ in the second integral $\Rightarrow -\sin xdx=du$
${\int }_0^{1/\sqrt{2}}\frac{dt}{t^4-t^2+1}+{\int }_1^{1/\sqrt{2}}\frac{-du}{u^4-u^2+1}$
${\int }_0^{1/\sqrt{2}}\frac{dt}{t^4-t^2+1}+{\int }_{1/\sqrt{2}}^1\frac{du}{u^4-u^2+1}$
$={\int }_0^{1/\sqrt{2}}\frac{dx}{x^4-x^2+1}+{\int }_{1/\sqrt{2}}^1\frac{dx}{x^4-x^2+1}$ (changing the variable)
$={\int }_0^1\frac{dx}{1-x^2+x^4}=I$