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What is #f(x) = int xsinx^2 + sec^2x dx# if #f(0)=-1 #?

Answer 1

Charles Arocha

#f(x)=-1/2cos(x^2)+tan(x)-1/2#

#f(x)=intxsin(x^2)dx+intsec^2(x)dx#

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

#f(x)=intxsin(x^2)dx+tan(x)+C#

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

#f(x)=1/2int2xsin(x^2)dx+tan(x)+C#

#f(x)=1/2intsin(u)du+tan(x)+C#

Since #intsin(x)dx=-cos(x)+C#:

#f(x)=-1/2cos(u)+tan(x)+C#

#f(x)=-1/2cos(x^2)+tan(x)+C#

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

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

#-1=-1/2(1)+0+C#

#C=-1/2#

Thus:

#f(x)=-1/2cos(x^2)+tan(x)-1/2#

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Answer 2

Christopher Agresta

The integral ( \int x\sin(x^2) + \sec^2(x) , dx ) cannot be directly integrated using elementary functions. Therefore, without additional information or techniques, it's not possible to determine ( f(x) ) and ( f(0) ) solely from the given information.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.