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integral dx/[secx+cosecx+tanx+cotx]^2 [] is normal bracket

integral dx/[secx+cosecx+tanx+cotx]^2
[] is normal bracket

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
I = \int \frac{1}{(secx+\csc x+tanx+cotx)^{2}}dx
Expand the integrand & put
u = tan(\frac{x}{2})
du = \frac{1}{2}sec^{2}(\frac{x}{2})dx
I = 2\int \frac{u^{2}(u-1)^{2}}{u^{6}+3u^{4}+3u^{2}+1}du
I = 2\int \frac{u^{2}(u-1)^{2}}{(u^{2}+1)^{3}}du
Now simply apply the partial fraction rule here & you will get
I = \frac{x}{2}-\frac{sinx}{2}+\frac{cosx}{2}-\frac{1}{8}cos2x+constant
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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