∫0π4sinx+cosxcos2x+sin4xdx=

I=∫0π4sinxdxcos2x+(1−cos2x)2+∫0π4cosxdx1−sin2x+sin4x t=cosx in the first integral and t=sinx in the second integral givesI=∫121dt1−t2+t4+∫012dt1−t2+t4 =∫01dt1−t2+t4 =∫01t−2t2+1t2−1 =12∫012(1t2)dtt2+1t2−1=12∫011+1t2−(1−1t2)t2+1t2−1dt =12∫011+1t2(t−1t)2+1dt−12∫011−1t2(t+1t)2−3dt =12tan −1t−1t01−143lnt+1t−3t+1t+301=π4+123ln(2+3)

∫0π4sinx+cosxcos2x+sin4xdx=

I=∫0π4sinxdxcos2x+(1−cos2x)2+∫0π4cosxdx1−sin2x+sin4x t=cosx in the first integral and t=sinx in the second integral givesI=∫121dt1−t2+t4+∫012dt1−t2+t4 =∫01dt1−t2+t4 =∫01t−2t2+1t2−1 =12∫012(1t2)dtt2+1t2−1=12∫011+1t2−(1−1t2)t2+1t2−1dt =12∫011+1t2(t−1t)2+1dt−12∫011−1t2(t+1t)2−3dt =12tan −1t−1t01−143lnt+1t−3t+1t+301=π4+123ln(2+3)

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$\int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{\cos^{2} x + \sin^{4} x} d x =$

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$\frac{π}{2} - \frac{1}{2 \sqrt{3}} \ln (2 + \sqrt{3})$

$\frac{π}{4} - \frac{1}{2 \sqrt{3}} \ln (2 + \sqrt{3})$

$\frac{π}{2} + \frac{1}{2 \sqrt{3}} \ln (2 + \sqrt{3})$

$\frac{π}{4} + \frac{1}{2 \sqrt{3}} \ln (2 + \sqrt{3})$

answer is D.

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Detailed Solution

$I = \int_{0}^{\frac{π}{4}} \frac{\sin x d x}{\cos^{2} x + {(1 - \cos^{2} x)}^{2}} + \int_{0}^{\frac{π}{4}} \frac{\cos x d x}{1 - \sin^{2} x + \sin^{4} x}$

$t = \cos x$ in the first integral and $t = \sin x$ in the second integral gives

$I = \int_{\frac{1}{\sqrt{2}}}^{1} \frac{d t}{1 - t^{2} + t^{4}} + \int_{0}^{\frac{1}{\sqrt{2}}} \frac{d t}{1 - t^{2} + t^{4}}$

$= \int_{0}^{1} \frac{d t}{1 - t^{2} + t^{4}}$ $= \int_{0}^{1} \frac{t^{- 2}}{t^{2} + \frac{1}{t^{2}} - 1}$

$= \frac{1}{2} \int_{0}^{1} \frac{2 (\frac{1}{t^{2}}) d t}{t^{2} + \frac{1}{t^{2}} - 1} = \frac{1}{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}} - (1 - \frac{1}{t^{2}})}{t^{2} + \frac{1}{t^{2}} - 1} d t$

$= \frac{1}{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 1} d t - \frac{1}{2} \int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 3} d t$

$= \frac{1}{2} {[\tan^{- 1} (t - \frac{1}{t})]}_{0}^{1} - \frac{1}{4 \sqrt{3}} \ln {[\frac{t + \frac{1}{t} - \sqrt{3}}{t + \frac{1}{t} + \sqrt{3}}]}_{0}^{1}$