indefinite integral
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Let y=x²+1, then dy=2xdx, so xdx=dy/2.

The integrand simplifies:

[(2sin(y)-sin(2y))/(sin(2y)+2sin(y))](dy/2)=

[(2sin(y)-2sin(y)cos(y))/(2sin(y)cos(y)+2sin(y)](dy/2)=

Cancel common factor sin(y):

(1-cos(y))/(2(1+cos(y)))dy=

Multiply by (1-cos(y))/(1-cos(y)):

(1-cos(y))²/(2(1-cos²(y)))dy=

(1-cos(y))²/(2sin²(y))dy=

(1-2cos(y)+cos²(y))/(2sin²(y))dy=

(csc²(y)/2-cot(y)csc(y)+cot²(y)/2)dy.

Derivative of cot(y) is -csc²(y).

Derivative of csc(y) is -cot(y)csc(y).

cot²(y)=csc²(y)-1.

∫(csc²(y)/2-cot(y)csc(y)+cot²(y)/2)dy=

∫(csc²(y)/2-cot(y)csc(y)+(csc²(y)-1)/2)dy=

∫(csc²(y)-cot(y)csc(y)-½)dy=

-cot(y)+csc(y)-y/2+c, where c is the constant of integration.

Replacing y with x²+1:

-cot(x²+1)+csc(x²+1)-(x²+1)/2+c or -cot(x²+1)+csc(x²+1)-x²+C (combining constants).

by Top Rated User (796k points)

Related questions

1 answer
asked May 2, 2019 in Calculus Answers by anonymous | 118 views
1 answer
1 answer
0 answers
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
85,650 questions
91,290 answers
2,205 comments
107,967 users