Question

solve the integral ill post to down

Answer 1

\[\int\limits_{}\sqrt{1 + sinx}*dx\]

Answer 2

We use a right triangle and Pythagoras theorem for substitutionse:

/|
/ |
/ |
/ |
1 / | u
/ |
/x |
--------

Now we can use the Pythagorean Theorem to find the third side, and
then the cosine of x. The third side is sqrt(1 - u^2), and so:

cos(x) = sqrt(1 - u^2)

Our integral is now:

sqrt(1 + u)
INT ------------- du
sqrt(1 - u^2)

Now I can't help but notice that

1 - u^2 = (1 - u)(1 + u)

so this fraction simplifies to:

1
INT ----------- du
sqrt(1 - u)

We are almost done. We have now transformed our trig integral into an
algebraic integral. Now a second substitution:

w = 1 - u

should finish the job.

If w = 1 - u, then dw = -du, so du = -dw. This gives:

1
- INT ------- dw
sqrt(w)

Interpret this as w^(-1/2), and we can use the formula for
antidifferentiating u^n:

1
- INT ------ dw = -2w^(1/2) + C
sqrt(w)

Now change back from w to u using w = 1 - u:

-2w^(1/2) = -2 sqrt(1 - u) + C

And now change back from u to x using u = sin(x):

-2w^(1/2) = -2 sqrt(1 - u) + C = -2 sqrt(1 - sin(x)) + C

I hope that helps.

Answer 3

really thanks

Answer 4

u r welcome....
do u think i hv done it right??????

Answer 5

well i have to check it...

Answer 6

ok, do so and let me know.........

Answer 7

^^