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Jairo Solis
Mathematics
26 May, 21:38
How to integrate sec^3x
+2
Answers (
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Rodney Villa
26 May, 21:41
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1 Use Integration by Parts on / int / sec^{3}x /, dx∫sec
3
xdx.
Let u=/sec{x}u=secx, dv=/sec^{2}xdv=sec
2
x, du=/sec{x}/tan{x} /, dxdu=secxtanxdx, v=/tan{x}v=tanx
2 Substitute the above into uv-/int v /, duuv-∫vdu.
/sec{x}/tan{x}-/int / tan^{2}x/sec{x} /, dxsecxtanx-∫tan
2
xsecxdx
3 Use Pythagorean Identities: / tan^{2}x=/sec^{2}x-1tan
2
x=sec
2
x-1.
/sec{x}/tan{x}-/int (/sec^{2}x-1) / sec{x} /, dxsecxtanx-∫ (sec
2
x-1) secxdx
4 Expand (/sec^{2}x-1) / sec{x} (sec
2
x-1) secx.
/sec{x}/tan{x}-/int / sec^{3}x-/sec{x} /, dxsecxtanx-∫sec
3
x-secxdx
5 Use Sum Rule: / int f (x) + g (x) /, dx=/int f (x) /, dx+/int g (x) /, dx∫f (x) + g (x) dx=∫f (x) dx+∫g (x) dx.
/sec{x}/tan{x}-/int / sec^{3}x /, dx+/int / sec{x} /, dxsecxtanx-∫sec
3
xdx+∫secxdx
6 Set it as equal to the original integral / int / sec^{3}x /, dx∫sec
3
xdx.
/int / sec^{3}x /, dx=/sec{x}/tan{x}-/int / sec^{3}x /, dx+/int / sec{x} /, dx∫sec
3
xdx=secxtanx-∫sec
3
xdx+∫secxdx
7 Add / int / sec^{3}x /, dx∫sec
3
xdx to both sides.
/int / sec^{3}x /, dx+/int / sec^{3}x /, dx=/sec{x}/tan{x}+/int / sec{x} /, dx∫sec
3
xdx+∫sec
3
xdx=secxtanx+∫secxdx
8 Simplify / int / sec^{3}x /, dx+/int / sec^{3}x /, dx∫sec
3
xdx+∫sec
3
xdx to 2/int / sec^{3}x /, dx2∫sec
3
xdx.
2/int / sec^{3}x /, dx=/sec{x}/tan{x}+/int / sec{x} /, dx2∫sec
3
xdx=secxtanx+∫secxdx
9 Divide both sides by 22.
/int / sec^{3}x /, dx=/frac{/sec{x}/tan{x}+/int / sec{x} /, dx}{2}∫sec
3
xdx=
2
secxtanx+∫secxdx
10 Original integral solved.
/frac{/sec{x}/tan{x}+/int / sec{x} /, dx}{2}
2
secxtanx+∫secxdx
11 Use Trigonometric Integration: the integral of / sec{x}secx is / ln{ (/sec{x}+/tan{x}) }ln (secx+tanx).
/frac{/sec{x}/tan{x}+/ln{ (/sec{x}+/tan{x}) }}{2}
2
secxtanx+ln (secx+tanx)
12 Add constant.
/frac{/sec{x}/tan{x}+/ln{ (/sec{x}+/tan{x}) }}{2}+C
2
secxtanx+ln (secx+tanx)
+ C
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