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Integrate arcsin x

∫ arcsin x dx

∫ arcsin x dx: To integrate arcsin x you can use this small trick by multiplying by 1 to make a product so that you can use the integration by parts formula to solve it. Multiplying by 1 does not change anything obviously but provides a means to use the standard parts formula. Hence ∫ arcsin x dx ≡ ∫ arcsin x × 1 dx

u Substitution

This time u=arcsin x and you can look up its derivative du/dx from the standard formula sheet if you cannot remember it, however this is straightforward.

This time we choose dv/dx to be 1 and therefore v=x. This is basic integration of a constant 1 which gives x.

Integration by parts.

Here we substitute the values of u, v, and du/dx into the integration by parts formula. If you cannot remember the formula then refer to the header on this page.

Here I have just rearranged the position of x to make it look neater. As you can see, a small part remains that still requires integration.

This part requires integration and is a simple one to do as you can use the u substitution method for it. How do you know when to use the u substitution method? As a rough rule if the derivative of u is anything remotely similar to x dx - - which is the outside bit -- then chances are it will work.

Here I am substituting and therefore u = 1 - x^2

The derivative of u is therefore -2x

We rearrange du/dx so that we get x dx on one side and everything else on the other side.

This might seem strange at first but it is all kosher :-) As you can see, we have an expression for x dx. Do not break open a bottle of Champaign just yet though because there is still some more work to do.

Here we can substitute for x dx and u = 1-x^2 to give an expression on the RHS which is essentially the same thing. I am keeping the terms in the same places so that you can see exactly what is happening.

I can simplify this a little by tucking away the ½ behind the integral sign to keep it out of the way.

In addition, the square root is the same as the power ½, and since this is a reciprocal, we include a minus sign to the power.

This is just a straight simple integration of u, and by adding 1 to the power the result is +½.

As you can see, the 2 cancels out.

Finally, we can now substitute the value for u back in.

We are almost near the result; however do not break open a bottle of Champaign just yet, as this requires substitution back into the original problem.

Here is the original problem. All you have to do is to substitute the contents of the square brackets, which we worked out in the previous step.

Here is the substitution, already made for you. The two negatives, inside and outside the brackets, obviously make a positive.

Final Result

Finally, a little bit of tidying and a final constant gives the result.