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Integral of dx/a+b*tan^2 x =?

1 Answer
Mar 19, 2018

Answer:

#1/((a/b-1)b)(x+sqrt(b/a)arctan(tanxsqrt(b/a))) + C#

Explanation:

#int(1/(a+btan^2x))dx#
put #tanx=t# #=>sec^2x dx=dt# #sec^2x=tan^2x+1#
#=>sec^2x=t^2+1#
#=>dx=dt/(t^2+1)#
#int(1/(a+bt^2))dt/(t^2+1)#
#int(1/(b(a/b+t^2)))dt/(t^2+1)#
multiply numerator and denominator with #a/b-1# we get
#int((a/b-1)/((a/b-1)(b(a/b+t^2)))dt/(t^2+1)#

add and subtract #t^2# in numerator we get
#int((t^2+a/b-1-t^2)/((a/b-1)(b(a/b+t^2)))dt/(t^2+1)#
#1/((a/b-1)b)int((t^2+a/b-1-t^2)/((a/b+t^2)(t^2+1)))dt#
#1/((a/b-1)b)int((-1)/((a/b+t^2))+1/(t^2+1)))dt#
#1/((a/b-1)b)(arctan(t)+sqrt(b/a)arctan(tsqrt(b/a)) + C#
substituting #t=tanx#
#1/((a/b-1)b)(arctan(tanx)+sqrt(b/a)arctan(tanxsqrt(b/a)) + C#
#arctan(tanx)=x#
#1/((a/b-1)b)(x+sqrt(b/a)arctan(tanxsqrt(b/a))) + C#