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Pure · Integration

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Integrating with trigonometric identities

Integrals like sin²x, cos²x and tan²x that have no direct rule — rewritten with the double-angle and Pythagorean identities into things you can integrate term by term.

Pure · Integration

Integrating with trigonometric identities

Integrals like sin²x, cos²x and tan²x that have no direct rule — rewritten with the double-angle and Pythagorean identities into things you can integrate term by term.

Why it works

There is no entry in the integral table for sin⁡2x\sin^2 x, and you cannot just "reverse a derivative" to find one — ∫sin⁡2x dx\int \sin^2 x\,dx is not 13sin⁡3x\tfrac13\sin^3 x (try differentiating that and you get sin⁡2xcos⁡x\sin^2 x\cos x, not sin⁡2x\sin^2 x). The way through is to rewrite the integrand using a trig identity until every term is something you can integrate.

Squares of sine and cosine come from rearranging the double-angle identities for cos⁡2x\cos 2x: cos⁡2x=1−2sin⁡2x  ⇒  sin⁡2x=12(1−cos⁡2x),\cos 2x = 1 - 2\sin^2 x \;\Rightarrow\; \sin^2 x = \tfrac12(1 - \cos 2x), cos⁡2x=2cos⁡2x−1  ⇒  cos⁡2x=12(1+cos⁡2x).\cos 2x = 2\cos^2 x - 1 \;\Rightarrow\; \cos^2 x = \tfrac12(1 + \cos 2x). The right-hand sides are just a constant plus a multiple of cos⁡2x\cos 2x — both standard integrals. So ∫sin⁡2x dx=∫12(1−cos⁡2x) dx=12x−14sin⁡2x+c.\int \sin^2 x\,dx = \int \tfrac12(1 - \cos 2x)\,dx = \tfrac12 x - \tfrac14\sin 2x + c.

tan⁡2x\tan^2 x uses the Pythagorean identity sec⁡2x=1+tan⁡2x\sec^2 x = 1 + \tan^2 x, so tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1, giving ∫tan⁡2x dx=∫(sec⁡2x−1) dx=tan⁡x−x+c.\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx = \tan x - x + c.

The general move is always the same: if you can't integrate it, reach for an identity that turns it into something you can — a double-angle identity to kill a square, a Pythagorean identity to swap tan⁡2\tan^2 for sec⁡2\sec^2, or a product-to-double- angle step such as 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x.