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Question Number 83146 by 09658867628 last updated on 28/Feb/20
∫π/20cotxcotx+tanxdx=
Answered by Kunal12588 last updated on 28/Feb/20
I=∫0π/2cotxcotx+tanxdx⇒I=∫0π/2tanxtanx+cotxdx⇒I=12∫0π/2dx⇒I=12×π2=π4∫π/20cotxcotx+tanxdx=π4
Answered by niroj last updated on 28/Feb/20
let,I=∫0π2cotxcotx+tanxdx.....(i)=∫0π2cot(π2−x)cot(π2−x)+tan(π2−x)dx∵∫0axdx=∫0a(a−x)dx=∫0π2tanxtanx+cotxdx......(ii)added(i)&(ii)2I=∫0π2(cotxcotx+tanx+tanxtanx+cotx)dx2I=∫0π2(cotx+tanxcotx+tanx)dx2I=∫0π2dx2I=[x]0π22I=(π2−0)2I=π2⇒I=π4//.
Commented by peter frank last updated on 28/Feb/20
thankyouboth
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