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Question: If I = \(\int_{}^{}\frac{dx}{\cos^{5}x + \sin^{5}x}\), then I equals...

If I = ∫dxcos⁡5x+sin⁡5x\int_{}^{}\frac{dx}{\cos^{5}x + \sin^{5}x}, then I equals

A

2\sqrt{2}log∣2+t2−t∣\left| \frac{\sqrt{2} + t}{\sqrt{2} - t} \right|+ 1(2−b2)b3\frac{1}{(2 - b^{2})b^{3}}log ∣b+tb−t∣\left| \frac{b + t}{b - t} \right| + 1(2−b2)2b3\frac{1}{(2 - b^{2})}\frac{2}{b^{3}}

tan–1 (tb)\left( \frac{t}{b} \right)+ C

B

5\sqrt{5}log∣5+t5−t∣\left| \frac{\sqrt{5} + t}{\sqrt{5} - t} \right|+ 12+b2\frac{1}{2 + b^{2}}log ∣b−tb+t∣\left| \frac{b - t}{b + t} \right| + 2b3\frac{2}{b^{3}}tan–1(tb)\left( \frac{t}{b} \right)+ C

C

5\sqrt{5}log∣5−t5+t∣\left| \frac{\sqrt{5} - t}{\sqrt{5} + t} \right|+ sin–1(tb)\left( \frac{t}{b} \right)+ C

D

5\sqrt{5}log∣2−t2+1∣\left| \frac{\sqrt{2} - t}{\sqrt{2} + 1} \right|+ sin–1(tb)\left( \frac{t}{b} \right)+ (2b3)\left( \frac{2}{b^{3}} \right)tan–1where t = sin x – cos x and b = 51/4(tb)\left( \frac{t}{b} \right)+ C

Answer

2\sqrt{2}log∣2+t2−t∣\left| \frac{\sqrt{2} + t}{\sqrt{2} - t} \right|+ 1(2−b2)b3\frac{1}{(2 - b^{2})b^{3}}log ∣b+tb−t∣\left| \frac{b + t}{b - t} \right| + 1(2−b2)2b3\frac{1}{(2 - b^{2})}\frac{2}{b^{3}}

tan–1 (tb)\left( \frac{t}{b} \right)+ C

Explanation

Solution

cos5 x + sin5 x = (cos x + sin x) [cos4 x –cos3x sin x + cos2 x sin2 x –cos x sin3 x + sin4 x]

= (cos x + sin x) [(cos2x + sin2x)2 – cos2 x sin2 x – cos x sin x (cos2 x + sin2 x)]

= (cos x + sin x) (1 –cos x sin x – cos2 x sin2 x)

\ I = ∫sin⁡x+cos⁡x(1+2sin⁡xcos⁡x)(1−cos⁡xsin⁡x–cos⁡2xsin⁡2x)\int_{}^{}\frac{\sin x + \cos x}{(1 + 2\sin x\cos x)(1 - \cos x\sin x–\cos^{2}x\sin^{2}x)}dx

Put sin x – cos x = t, so that (cos x + sin x) dx = dt, and t2 = 1 –2 sin x cos x

Thus

I = ∫dt(2−t2)[1−t2−12−14(t2−1)2]\int_{}^{}\frac{dt}{(2 - t^{2})\left\lbrack 1 - \frac{t^{2} - 1}{2} - \frac{1}{4}(t^{2} - 1)^{2} \right\rbrack}

= 4

4 dt

= 2\sqrt { 2 }log ∣2+t2−t∣\left| \frac{\sqrt{2} + t}{\sqrt{2} - t} \right| 12−5\frac { 1 } { 2 - \sqrt { 5 } } 153/4\frac{1}{5^{3/4}}log ∣51/4+t51/4−t∣\left| \frac{5^{1/4} + t}{5^{1/4} - t} \right| + 12+5\frac { 1 } { 2 + \sqrt { 5 } } 253/4\frac{2}{5^{3/4}}

tan–1 (t51/4)\left( \frac{t}{5^{1/4}} \right)+ C

where t = sin x – cos x