What is f(x) = int xsinx^2 + sec^2x dxf(x)=∫xsinx2+sec2xdx if f(0)=-1 f(0)=−1?

1 Answer
mason m
Sep 25, 2016

f(x)=-1/2cos(x^2)+tan(x)-1/2f(x)=−12cos(x2)+tan(x)−12

Explanation:

f(x)=intxsin(x^2)dx+intsec^2(x)dxf(x)=∫xsin(x2)dx+∫sec2(x)dx

The second integral is simple: since d/dxtan(x)=sec^2(x)ddxtan(x)=sec2(x), we see that intsec^2(x)dx=tan(x)+C∫sec2(x)dx=tan(x)+C.

f(x)=intxsin(x^2)dx+tan(x)+Cf(x)=∫xsin(x2)dx+tan(x)+C

For the remaining integral, use the substitution u=x^2u=x2. This implies that du=2xdxdu=2xdx.

f(x)=1/2int2xsin(x^2)dx+tan(x)+Cf(x)=12∫2xsin(x2)dx+tan(x)+C

f(x)=1/2intsin(u)du+tan(x)+Cf(x)=12∫sin(u)du+tan(x)+C

Since intsin(x)dx=-cos(x)+C∫sin(x)dx=−cos(x)+C:

f(x)=-1/2cos(u)+tan(x)+Cf(x)=−12cos(u)+tan(x)+C

f(x)=-1/2cos(x^2)+tan(x)+Cf(x)=−12cos(x2)+tan(x)+C

Using the initial condition f(0)=-1f(0)=−1, we see that:

-1=-1/2cos(0)+tan(0)+C−1=−12cos(0)+tan(0)+C

-1=-1/2(1)+0+C−1=−12(1)+0+C

C=-1/2C=−12

Thus:

f(x)=-1/2cos(x^2)+tan(x)-1/2f(x)=−12cos(x2)+tan(x)−12

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