Calculus · real student question

Find the indefinite integral of (1 + tan^2 y) with respect to y.

Question

Find the indefinite integral

(1+tan2y)dy.\int\left(1+\tan^{2}y\right)dy.

Step-by-step solution

  1. Do not integrate term by term. Splitting into 1dy+tan2ydy\int 1\,dy+\int\tan^{2}y\,dy is legal but pointless: the second piece has no elementary antiderivative you can write down without first using the very identity that solves the whole problem.

  2. Apply the Pythagorean identity. Starting from sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1 and dividing every term by cos2θ\cos^2\theta gives

    tan2y+1=sec2y.\tan^{2}y+1=\sec^{2}y.

    So the integrand is simply sec2y\sec^{2}y.

  3. Recognise the derivative you already know. Since

    ddytany=sec2y,\frac{d}{dy}\tan y=\sec^{2}y,

    the integral is immediate:

    sec2ydy=tany+C.\int\sec^{2}y\,dy=\tan y+C.

  4. State the result with its constant of integration.

    (1+tan2y)dy=tany+C.\int\left(1+\tan^{2}y\right)dy=\tan y+C.

  5. Verify by differentiating back. Using the quotient rule on tany=sinycosy\tan y=\dfrac{\sin y}{\cos y}:

    ddytany=cos2y+sin2ycos2y=1cos2y=sec2y=1+tan2y,\frac{d}{dy}\tan y=\frac{\cos^{2}y+\sin^{2}y}{\cos^{2}y}=\frac{1}{\cos^{2}y}=\sec^{2}y=1+\tan^{2}y,

    which is the original integrand exactly. Note the antiderivative is valid on each interval where cosy0\cos y\neq 0, that is between consecutive odd multiples of π2\tfrac{\pi}{2}.

Answer

tany+C\tan y+C

Need to solve a different problem like this? Open the solver →