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int_(-(pi)/(6))^((pi)/(3))(-sin x)/(cos^(2)x)dx

π6π3sinxcos2xdx \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{-\sin x}{\cos ^{2} x} d x

Full solution

Q. π6π3sinxcos2xdx \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{-\sin x}{\cos ^{2} x} d x
  1. Recognize integral type: Step 11: Recognize the integral type.\newlineWe have the integral of a trigonometric function, which suggests a potential substitution. The integral is π6π3sinxcos2xdx\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{-\sin x}{\cos^2 x} \, dx. Recognizing that ddx(tanx)=sec2x\frac{d}{dx}(\tan x) = \sec^2 x, we can use a substitution.
  2. Perform substitution: Step 22: Perform the substitution.\newlineLet u=tanx u = \tan x , then du=sec2xdx du = \sec^2 x \, dx . Since sec2x=1/cos2x\sec^2 x = 1/\cos^2 x, we can rewrite dx dx as dx=cos2xdu dx = \cos^2 x \, du . Substituting into the integral, we get:\newlinesinxcos2xdx=sinxdu \int \frac{-\sin x}{\cos^2 x} \, dx = \int -\sin x \, du \newlineThis is incorrect because we didn't adjust the sinx\sin x term properly for the substitution.

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