Author Topic: A few qstns (Read 1271 times) Tweet Share

perfectscore · « **on:** July 11, 2008, 03:12:59 pm »

Please help me out thx.
the f's are integration signs

Show full working.

f sec(x)^3.tanx dx

f cosx/(3+2sinx)^2 dx

f tan(3x)^2 dx

Thanks :]

dcc · « **Reply #1 on:** July 11, 2008, 03:54:41 pm »

Q1:

$I = \int \sec^3 x \cdot \tan x \; dx = \int \dfrac{1}{\cos^3 x} \cdot \dfrac{\sin x}{\cos x}\; dx$

$I = \int \dfrac{\sin x}{\cos^4 x}\; dx$

Let $u = \cos x$

$I = - \int \dfrac{1}{u^4}\; du = \dfrac{1}{3} u^{-3} + C$

$I = \dfrac{1}{3} \sec^3 x + C$

Q2:

$I = \int \dfrac{\cos x}{\left(3 + 2 \sin x\right)^2}\; dx$

Let $u = 3 + 2\sin x \implies \dfrac{du}{dx} = 2\cos x$

$I = \dfrac{1}{2} \int \dfrac{1}{u^2}\; du = -\dfrac{1}{2} \cdot \dfrac{1}{u} + C$

$I = -\dfrac{1}{6 + 4\sin x} + C$

Q3:

$\int \tan^2 3x\; dx = \int \sec^2 3x - 1 \; dx = \dfrac{1}{3} \tan 3x - x + c$

perfectscore · « **Reply #2 on:** July 23, 2008, 10:50:23 pm »

What about:

f (sinx)^2/(cosx)^2 dx

and

f sinx(secx^2) dx

phagist_ · « **Reply #3 on:** July 23, 2008, 11:02:51 pm »

$\int \frac{sin^2x}{cos^2x} ~dx$

= $\int tan^2x~ dx$

Given $Q3$ above, you should be able to figure it out

.

edit.. is that $sec^2x$ or $secx^2$ ?

perfectscore · « **Reply #4 on:** July 24, 2008, 07:01:19 pm »

sec squared x

dcc · « **Reply #5 on:** July 24, 2008, 07:08:06 pm »

Quote from: perfectscore on July 23, 2008, 10:50:23 pm

$\int \sin x \left( \sec^2 x \right)\ dx$

$I = \int \sin x \left( \dfrac{1}{\cos^2 x} \right)\ dx$

Let $t = \cos x \implies \dfrac{dt}{dx} = - \sin x$

$I = \int \left(- \dfrac{dt}{dx} \right) \left( \dfrac{1}{t^2} \right)\ dx = - \int \dfrac{1}{t^2}\ dt$

$I = \dfrac{1}{t} + C = \dfrac{1}{\cos x} + C$

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A few qstns

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