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        sinx.log(sinx)dx=log2-1Limit of integration:0 to pi/2
one year ago

Saurabh Koranglekar
3160 Points
							Dear studentPlease ask a question

one year ago
Phanindra
226 Points
							Put sinx = t... (1)then limits also will change...Lower limit = 0 and upper limit = 1 Now.. differentiate (1) that is cosxdx=dt  so..dx = dt/cosx = $dt/\sqrt{}(1-t^2)$… Because ...sin2x + cos2x = 1...cosx = $\sqrt{}(1-t^2)$…Now the question chages and its limits are...0 and 1...Hope you have understood the steps...This is the procedure...You will get the answer..If you do not get the answer..then tell me.i will send the full solution..

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

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