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Question Number 123488 by peter frank last updated on 25/Nov/20
Answered by MJS_new last updated on 25/Nov/20
x=sechy=1coshy=2eye2y+1⇒y=ln1+1−x2x=sech−1x∫sech−1xdx=∫ln1+1−x2xdx=[byparts]=xln1+1−x2x+∫dx1−x2==xln1+1−x2x+arcsinx+C[=xsech−1x+arcsinx+C]
Answered by Dwaipayan Shikari last updated on 26/Nov/20
sech−1x=tx=secht⇒2et+e−t⇒et+e−t=2x⇒e2t−2etx+1=0et=1x+1−x2x⇒t=log(1+1−x2x)∫log(1+1−x2x)dx=∫log(1+1−x2)−∫logxdx=xlog(1+1−x2)+∫x21−x21+1−x2dx−xlogx+x=xlog(1+1−x2x)+∫11−x2dx=xlog(1+1−x2x)+sin−1x+C
Commented by peter frank last updated on 26/Nov/20
thankyou
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