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∫{[sin(x+α)/cos 3 x][√[(cosecx + secx)/(cosecx-secx)]}dx

∫{[sin(x+α)/cos3x][√[(cosecx + secx)/(cosecx-secx)]}dx

Grade:12th Pass

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
I = \int \frac{sin(x+\alpha )}{cos^{3}(x)}.\sqrt{\frac{csc(x)+sec(x)}{csc(x)-sec(x)}}dx
I = \int \frac{sin(x+\alpha )}{cos^{3}(x)}.\sqrt{\frac{cos(x)+sin(x)}{cos(x)-sin(x)}}dx
I = \int \frac{sin(x+\alpha )}{cos^{3}(x)}.\sqrt{\frac{1+tan(x)}{1-tan(x)}}dx
I = \int \frac{sin(x)cos\alpha+cos(x)sin\alpha }{cos(x)}.\sqrt{\frac{1+tan(x)}{1-tan(x)}}.sec^{2}(x)dx
t = tan(x)
dt = sec^{2}(x).dx
I = \int (tcos\alpha +sin\alpha )\sqrt{\frac{1+t}{1-t}} dt
I = cos\alpha \int t\sqrt{\frac{1+t}{1-t}}dt + sin\alpha \int \sqrt{\frac{1+t}{1-t}}dt
I = cos\alpha \int \frac{t(t+1)}{\sqrt{1-t^{2}}}dt + sin\alpha \int \frac{t+1}{\sqrt{1-t^{2}}}dt
Simply apply the substitution method here,
I = \frac{cos\alpha }{2}.(sin^{-1}(t)-(t+2)\sqrt{1-t^{2}})+sin\alpha (sin^{-1}(t)-\sqrt{1-t^{2}}) + constant
I = \frac{cos\alpha }{2}.(sin^{-1}(tanx)-(tanx+2)\sqrt{1-tanx^{2}})+sin\alpha (sin^{-1}(tanx)-\sqrt{1-tanx^{2}}) + constant
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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