No basic rule integrates \sin^2 x directly. The power rule doesn't apply to a trig function raised to a power, and nothing about \sin^2 x looks like a chain-rule derivative waiting to be undone. The fix is entirely algebraic: rewrite the integrand using a trig identity FIRST, into a form the existing rules already handle. That's the same "reshape before integrating" habit the previous section built for algebraic expressions.
\sin^2 x and \cos^2 x each oscillate too "busily" for the basic rules to see a pattern in. Rewritten as a shifted, scaled copy of \cos 2x plus a constant, that same oscillation becomes the familiar, already-integrable kind. The identity doesn't change what the function IS, only what it looks like. It's the same function wearing algebra the existing toolkit can actually parse.
A trig identity used for integration is an algebraic rewrite of a trig expression into an equivalent but more integrable form. The identity doesn't change what the function IS, only what it looks like. The half-angle identities \sin^2 x = \frac{1-\cos 2x}{2} and \cos^2 x = \frac{1+\cos 2x}{2} turn an un-integrable squared trig function into a sum of two pieces that both integrate immediately: a constant and a cosine of a linear argument. This is the single most-used trick the full trigonometric-integrals chapter, several chapters ahead, builds an entire technique out of. It's worth being automatic with now, on the simplest possible integrands. That chapter compounds it later with products of several trig functions at once.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Which expression equals \dfrac{\sin\theta}{\cos\theta} by definition?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
cotθ is the flipped ratio, cosθ/sinθ — this ratio (sine over cosine) is tanθ.
secθ is a reciprocal identity (1/cosθ), not a ratio of sine and cosine.
cscθ is a reciprocal identity (1/sinθ), not a ratio of sine and cosine.
Why
tanθ = sinθ/cosθ is the quotient identity — sine on top, cosine on bottom.
From College Algebra
Which expression is equivalent to \cot\theta using a reciprocal identity?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
That's cscθ — the reciprocal of sinθ, not of tanθ.
That's secθ — the reciprocal of cosθ, not of tanθ.
tanθ is cotθ's reciprocal PARTNER, not cotθ itself — flipping tanθ gives cotθ, they aren't the same expression.
Why
cotθ is defined as the reciprocal of tanθ: cotθ = 1/tanθ.
Confirm the half-angle rewrite live
Drag the probe across the domain -- both sides agree everywhere, confirming the identity ahead of the integration worked example below.
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The matching identity for cosine-squared
Same check, for the cosine version this section's practice bank also uses.
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Worked example
Integrating sin-squared with the half-angle identity
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1
Evaluate \displaystyle\int \sin^2 x\,dx.
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2
Rewrite using the half-angle identity: \sin^2 x = \dfrac{1-\cos 2x}{2}.
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3the inner 2x needs its own factor of 1/2 from a quick u-substitution
Split and integrate each piece: \int \dfrac{1}{2}\,dx - \int \dfrac{\cos 2x}{2}\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4}.
the inner 2x needs its own factor of 1/2 from a quick u-substitution
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4
\boxed{\int \sin^2 x\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C}
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Where this goes next
Trigonometric integrals, several chapters ahead
This half-angle move is a single case of a much larger technique: integrating products and powers of trig functions. That technique gets a whole chapter to itself once techniques of integration begin in earnest.
Check your understanding (counts toward your progress)
Find \displaystyle\int_0^{\pi}551\sin^2(x)\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\displaystyle\int_0^{\pi}551\cdot\dfrac{1-\cos(2x)}{2}\,dx=\dfrac{551}{2}\left[x-\dfrac{\sin(2x)}{2}\right]_0^{\pi}=\dfrac{551\pi}{2}\approx865.5088.
Correct answer
Also accepted:
Why
Use the Pythagorean-derived identity \sin^2(x)=\dfrac{1-\cos(2x)}{2} before integrating.
Find \displaystyle\int \cos^2 x\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Splitting gives \dfrac{x}{2} + \dfrac{\sin 2x}{4}, plus C.
Correct answer
Also accepted:
Why
Use \cos^2 x = \dfrac{1+\cos 2x}{2}.
Find \displaystyle\int_0^{\pi}476\sin^2(x)\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\displaystyle\int_0^{\pi}476\cdot\dfrac{1-\cos(2x)}{2}\,dx=\dfrac{476}{2}\left[x-\dfrac{\sin(2x)}{2}\right]_0^{\pi}=\dfrac{476\pi}{2}\approx747.6991.
Correct answer
Also accepted:
Why
Use the Pythagorean-derived identity \sin^2(x)=\dfrac{1-\cos(2x)}{2} before integrating.