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Viktor [21]
4 years ago
9

How to solve this indefinite integral? ​

Mathematics
1 answer:
Karolina [17]4 years ago
3 0

\displaystyle\int\frac{3x}{\cos^2(2x^2)}\,\mathrm dx

Substitute y=2x^2, so that \mathrm dy=4x\,\mathrm dx:

\displaystyle\int\frac{3x}{\cos^2(2x^2)}\,\mathrm dx=\frac34\int\frac{4x}{\cos^2(2x^2)}\,\mathrm dx=\frac34\int\frac{\mathrm dy}{\cos^2y}

Then

\dfrac1{\cos^2y}=\sec^2y=\dfrac{\mathrm d}{\mathrm dy}[\tan y]

so that the integral wrt y comes out to be

\displaystyle\frac34\int\sec^2y\,\mathrm dy=\frac34\tan y+C

Replace y to solve for the integral wrt x:

\displaystyle\int\frac{3x}{\cos^2(2x^2)}\,\mathrm dx=\boxed{\frac34\tan(2x^2)+C}

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