If dx/³ x + ³ x = A ≤ft| {{2}+t}{{2}-t} | + B ^{-1}(t) + c, then ≤ft( B/A, t ) = [AP EAPCET 21 May 2025 S1]

Correct Answer is : (22,sin⁡x−cos⁡x)(2{2}, x - x)(22,sinx−cosx)

Correct Answer is : (22,sin⁡x−cos⁡x)(2{2}, x - x)(22,sinx−cosx)

Test Index

Indefinite Integration

Section: Mathematics

Question : 42 of 89

Marks: +1, -0

\int \frac{dx}{\sin^3 x + \cos^3 x} = A \log \left| \frac{\sqrt{2}+t}{\sqrt{2}-t} \right| + B \tan^{-1}(t) + c

\left( \frac{B}{A}, t \right) =

Solution:

\text{ Let } I = \int \frac{1}{\sin^3 x + \cos^3 x} \, dx

= \int \frac{1}{(\sin x + \cos x)(\sin^2 x + \cos^2 x - \sin x \cos x)} \, dx

= \int \frac{\sin x + \cos x}{(\sin x + \cos x)^2 (1 - \sin x \cos x)} \, dx

= \int \frac{\sin x + \cos x}{(1 + 2 \sin x \cos x)(1 - \sin x \cos x)} \, dx

= \int \frac{dt}{(1+1-t^2)\left(1-\frac{1-t^2}{2}\right)}

Put

\sin x - \cos x = t

(\cos x + \sin x) \, dx = dt

1 - 2 \sin x \cos x = t^2

2 \sin x \cos x = 1 - t^2

= \int \frac{2 \, dt}{(2-t^2)(1+t^2)} = -2 \int \frac{dt}{(t^2+1)(t^2-2)}

= -\frac{2}{3} \int \left( \frac{1}{t^2-2} - \frac{1}{t^2+1} \right) dt

= -\frac{2}{3} \left[ \frac{1}{2\sqrt{2}} \ln \left( \frac{t-\sqrt{2}}{t+\sqrt{2}} \right) - \tan^{-1} t \right]

= -\frac{1}{3\sqrt{2}} \ln \left( \frac{\sin x - \cos x - \sqrt{2}}{\sin x - \cos x + \sqrt{2}} \right) + \frac{2}{3} \tan^{-1}(\sin x - \cos x) + C

= \frac{1}{3\sqrt{2}} \ln \left| \frac{\sin x - \cos + \sqrt{2}}{\sin x - \cos x - \sqrt{2}} \right| + \frac{2}{3} \tan^{-1}(\sin x - \cos x) + C

\because A = \frac{1}{3\sqrt{2}}, B = \frac{2}{3} \text{ and } t = \sin x - \cos x

\because \left( \frac{B}{A}, t \right) = (2\sqrt{2}, \sin x - \cos x)

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