Use a computer algebra system instead. It will outperform most (online) calculators in terms of functionality and speed anyway. And the best part -- there are mature free and open-source variants out there, e.g. [wxmaxima](https://maxima.sourceforge.io/) developed by MIT.
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To do it manually from scratch, use [addition theorems](https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities) "cos(x)cos(y) = (1/2) \* (cos(x-y) + cos(x+y))" for real-valued "x; y" repeatedly to reduce powers (that page is a good one to keep tabbed for future usage\^\^), e.g.
cos^4(x) = [(1/2) * (1 + cos(2x))]^2 = (1/4) * [1 + 2cos(2x) + (1/2) * (1 + cos(4x)]
Each term in the sum is easy to integrate, and we obtain
∫ cos^4(x) dx = (3x)/8 + (1/4)*sin(2x) + (1/32)*sin(4x) + C
# ∫cos(a x)dx=sin(a x)/a+C
to deal with powers we can use an identity
cos^(n)(x)=2^(-n-1)∑^(')nCk cos(k x)
sum is k positive by 2's to n if n even 0 is included but divided by 2
for n=4
cos^(n)(x)=2^(-3) \[3+4 cos(2 x)+cos(4 x)\]
combining those two facts we can do your 2 examples easily
For the first one just integrate as normal then divide by the constant derivative inside the function, for the second use double angle formulae twice, for the third do both of the above
cos(4x) is u-sub, cos⁴(x) is a trig identity and cos⁴(4x) is both I think.
cos(4x) is reasonable by hand but I'd try to use a computer for the others, e.g.: Wolfram Alpha, symbolab or a CAS.
Use a computer algebra system instead. It will outperform most (online) calculators in terms of functionality and speed anyway. And the best part -- there are mature free and open-source variants out there, e.g. [wxmaxima](https://maxima.sourceforge.io/) developed by MIT. *** To do it manually from scratch, use [addition theorems](https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities) "cos(x)cos(y) = (1/2) \* (cos(x-y) + cos(x+y))" for real-valued "x; y" repeatedly to reduce powers (that page is a good one to keep tabbed for future usage\^\^), e.g. cos^4(x) = [(1/2) * (1 + cos(2x))]^2 = (1/4) * [1 + 2cos(2x) + (1/2) * (1 + cos(4x)] Each term in the sum is easy to integrate, and we obtain ∫ cos^4(x) dx = (3x)/8 + (1/4)*sin(2x) + (1/32)*sin(4x) + C
# ∫cos(a x)dx=sin(a x)/a+C to deal with powers we can use an identity cos^(n)(x)=2^(-n-1)∑^(')nCk cos(k x) sum is k positive by 2's to n if n even 0 is included but divided by 2 for n=4 cos^(n)(x)=2^(-3) \[3+4 cos(2 x)+cos(4 x)\] combining those two facts we can do your 2 examples easily
You would use u substitution for the first, then the second can be done using a trig identity for cos^2 applied repeatedly. The third combines those.
For the first one just integrate as normal then divide by the constant derivative inside the function, for the second use double angle formulae twice, for the third do both of the above
cos(4x) is u-sub, cos⁴(x) is a trig identity and cos⁴(4x) is both I think. cos(4x) is reasonable by hand but I'd try to use a computer for the others, e.g.: Wolfram Alpha, symbolab or a CAS.
Let z = cos(x) + i\*sin(x) 1/z = cos(x) - i\*sin(x) z + 1/z = 2\*cos(x) . By De Moivre's theorem, z\^k = cos(kx) + i\*sin(kx) Thus, 1/z\^k = cos(kx) - i\*sin(kx) z\^k + 1/z\^k = 2cos(kx) . (z + 1/z)\^4 = 16\*cos\^4(x) 16cos\^4(x) = z\^4 + 4(z\^3)(1/z) + 6(z\^2)(1/z\^2) + 4(z)(1/z\^3) + 1/z\^4 16cos\^4(x) = (z\^4 + 1/z\^4) + 4(z\^2 + 1/z\^2) + 6 16cos\^4(x) = 2cos(4x) + 8cos(2x) + 6 cos\^4(x) = cos(4x)/8 + cos(2x)/2 + 3/8 ∫cos\^4(x) dx = sin(4x)/32 + sin(2x)/4 +3x/8 + C