Problem to solve
Find the indefinite integral \[\int\frac{3\sin^4[x]-2\cos^2[x]}{1+\sin^2[x]} dx\]
Solution
STEP 1
Take the integral: integral(3 sin^4(x) - 2 cos^2(x))/(sin^2(x) + 1) dx
STEP 2
Multiply numerator and denominator of (3 sin^4(x) - 2 cos^2(x))/(sin^2(x) + 1) by csc^6(x): = integral(3 csc^2(x) - 2 cot^2(x) csc^4(x))/(csc^6(x) + csc^4(x)) dx
STEP 3
Prepare to substitute u = cot(x). Rewrite (3 csc^2(x) - 2 cot^2(x) csc^4(x))/(csc^6(x) + csc^4(x)) using csc^2(x) = cot^2(x) + 1: = integral-((2 cot^4(x) + 2 cot^2(x) - 3) csc^2(x))/((cot^2(x) + 1)^2 (cot^2(x) + 2)) dx
STEP 4
For the integrand -((2 cot^4(x) + 2 cot^2(x) - 3) csc^2(x))/((cot^2(x) + 1)^2 (cot^2(x) + 2)), substitute u = cot(x) and du = -csc^2(x) dx: = integral(2 u^4 + 2 u^2 - 3)/((u^2 + 1)^2 (u^2 + 2)) du
STEP 5
For the integrand (2 u^4 + 2 u^2 - 3)/((u^2 + 1)^2 (u^2 + 2)), use partial fractions: = integral1/(u^2 + 2) + 1/(u^2 + 1) - 3/(u^2 + 1)^2 du
STEP 6
Integrate the sum term by term and factor out constants: = integral1/(u^2 + 2) du + integral1/(u^2 + 1) du - 3 integral1/(u^2 + 1)^2 du
STEP 7
Factor 2 from the denominator: = integral1/(2 (u^2/2 + 1)) du + integral1/(u^2 + 1) du - 3 integral1/(u^2 + 1)^2 du
STEP 8
Factor out constants: = 1/2 integral1/(u^2/2 + 1) du + integral1/(u^2 + 1) du - 3 integral1/(u^2 + 1)^2 du
STEP 9
For the integrand 1/(u^2/2 + 1), substitute s = u/sqrt(2) and ds = 1/sqrt(2) du: = 1/sqrt(2) integral1/(s^2 + 1) ds + integral1/(u^2 + 1) du - 3 integral1/(u^2 + 1)^2 du
STEP 10
The integral of 1/(s^2 + 1) is tan^(-1)(s): = (tan^(-1)(s))/sqrt(2) + integral1/(u^2 + 1) du - 3 integral1/(u^2 + 1)^2 du
STEP 11
The integral of 1/(u^2 + 1) is tan^(-1)(u): = (tan^(-1)(s))/sqrt(2) + tan^(-1)(u) - 3 integral1/(u^2 + 1)^2 du
STEP 12
For the integrand 1/(u^2 + 1)^2, substitute u = tan(p) and du = sec^2(p) dp. Then (u^2 + 1)^2 = (tan^2(p) + 1)^2 = sec^4(p) and p = tan^(-1)(u): = (tan^(-1)(s))/sqrt(2) + tan^(-1)(u) - 3 integral cos^2(p) dp
STEP 13
Write cos^2(p) as 1/2 cos(2 p) + 1/2: = (tan^(-1)(s))/sqrt(2) + tan^(-1)(u) - 3 integral(1/2 cos(2 p) + 1/2) dp
STEP 14
Integrate the sum term by term and factor out constants: = (tan^(-1)(s))/sqrt(2) + tan^(-1)(u) - 3/2 integral cos(2 p) dp - 3/2 integral1 dp
STEP 15
For the integrand cos(2 p), substitute w = 2 p and dw = 2 dp: = (tan^(-1)(s))/sqrt(2) + tan^(-1)(u) - 3/4 integral cos(w) dw - 3/2 integral1 dp
STEP 16
The integral of cos(w) is sin(w): = (tan^(-1)(s))/sqrt(2) + tan^(-1)(u) - (3 sin(w))/4 - 3/2 integral1 dp
STEP 17
The integral of 1 is p: = -(3 p)/2 + (tan^(-1)(s))/sqrt(2) + tan^(-1)(u) - (3 sin(w))/4 + constant
STEP 18
Substitute back for w = 2 p: = -(3 p)/2 - 3/4 sin(2 p) + (tan^(-1)(s))/sqrt(2) + tan^(-1)(u) + constant
STEP 19
Substitute back for p = tan^(-1)(u): = 1/2 (sqrt(2) tan^(-1)(s) - (3 u)/(u^2 + 1) - tan^(-1)(u)) + constant
STEP 20
Substitute back for s = u/sqrt(2): = 1/2 (-(3 u)/(u^2 + 1) - tan^(-1)(u) + sqrt(2) tan^(-1)(u/sqrt(2))) + constant
STEP 21
Substitute back for u = cot(x): = 1/2 (-3 sin(x) cos(x) - tan^(-1)(cot(x)) + sqrt(2) tan^(-1)(cot(x)/sqrt(2))) + constant
STEP 22
An alternative form of the integral is: = 1/4 (-3 sin(2 x) - 2 tan^(-1)(cot(x)) + 2 sqrt(2) tan^(-1)(cot(x)/sqrt(2))) + constant
STEP 23
Which is equivalent for restricted x values to: Answer: | | = x/2 - 3/4 sin(2 x) - (tan^(-1)(sqrt(2) tan(x)))/sqrt(2) + constant
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Edited by tododemate® - 2021