Need help with evaluating integrals please?

Question

evaluate these integrals

1) (7+15 cos^2 x) / (cos^2 x) dx

b=pi / 4 a=0

2) 2 / (square root of (1-x^2)) dx

b=1 a=sqrt2 / 2

Answer 1

1) ∫ [(7 + 15 cos²x) / cos²x] dx =

break it up into:

∫ [(7 / cos²x) + (15 cos²x / cos²x)] dx =

∫ (7 / cos²x) dx  + ∫ (15 cos²x / cos²x) dx =

that simplifies into:

7 ∫ (1 / cos²x) dx  + 15 ∫ dx =

7 tanx + 15x + C

then evaluate the definte integral, that is:

[7 tan(pi/4) + 15(pi/4)] - [7 tan(0) + 15(0)] =

7(1) + (15/4)pi - 0 =

7 + (15/4)pi

2) ∫ [2 / √(1 - x²)] dx =

take the constant out:

2 ∫ [1 / √(1 - x²)] dx =

that is an immediate integral:

2 arcsinx + C

then, plugging in the integration limit, note that the integrand doesn't exist at x = 1, thus this is an improper integral, indeed, and therefore you have to evaluate the limit:

lim 2 arcsinx - 2 arcsin[(√2)/2] = 2 (pi/2) - 2 (pi/4) = pi - (pi/2) = pi/2

x → 1(-)

your integral converges to pi/2

I hope it helps...

Bye!

Answer 2

∫ ( 7 + 15 cos²x ) / ( cos²x ) dx

Split the fraction:

= ∫ [ ( 7 ) / ( cos²x ) + ( 15 cos²x ) / ( cos²x ) ] dx

Cancel out cosines and note that 1/cos(x) = sec(x)

= ∫ ( 7·sec²x + 15 ) dx

The derivative of tan(x) is sec²(x), so the integral is easy:

= [ 7·tan(x) + 15·x ] as x goes from 0 to π/4

= [ 7·tan(π/4) + 15·π/4 ] - [ 7·tan(0) + 15·0 ]

= [ 7·(1) + 15·π/4 ] - [ 0 + 0 ]

= 7 + 15·π/4

——————————————————————————————————————

∫ 2 / √( 1 - x² ) dx

Make a trigonometric substitution:

Let sin(u) = x

Then cos(u)du = dx

→ ∫ 2 / √[ 1 - sin²(u) ] cos(u)du

Use the Pythagorean trigonometric identity:

sin²θ + cos²θ = 1

cos²θ = 1 - sin²θ

→ ∫ 2 / √[ cos²(u) ] cos(u)du

= ∫ 2 / | cos(u) | cos(u)du

Cosine is positive over this interval, so absolute value can be lost.

The cosines cancel out:

= ∫ 2 du

= [ 2·u ] as x goes from √(2)/2 to 1

Since sin(u) = x

Then u = arcsin(x)

→ [ 2·arcsin(x) ] as x goes from √(2)/2 to 1

Note that at x = 1, the original integrant is undefined, as it has 0 in the denominator. That makes this technically an "improper integral". Thus, you would take the limit as x approaches 1 from the left for the upper bound:

= {{{ lim x → 1¯ }}} [ 2·arcsin(x) ] - [ 2·arcsin( √(2)/2 ) ]

As arcsine is defined when x = 1, you can just use that value in arcsine. It is basically just a formality here.

= [ 2·arcsin(1) ] - [ 2·arcsin( √(2)/2 ) ]

= [ 2·π/2 ] - [ 2·π/4 ]

= [ π ] - [ π/2 ]

= π/2

Note that for this one, 1/√( 1 - x² ) is the derivative of arcsin(x), so you can immediately get the indefinite integral 2·arcsin(x) it if you recall that.