Question

Verify the following reduction formula:
$\displaystyle \int \sec^n(u)\, du=\frac{\sec^{n-2}(u)\tan(u)}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2}(u)\, du, \; n\neq 1$​

Answer 1

Let,

$\displaystyle\longrightarrow I=\int\sec^nu\ du$

Taking $n=n-2+2,$

$\displaystyle\longrightarrow I=\int\sec^{n-2+2}u\ du$

$\displaystyle\longrightarrow I=\int\sec^{n-2}u\cdot \sec^2u\ du$

Performing integration by parts,

$\displaystyle\longrightarrow I=\sec^{n-2}u\tan u-\int\dfrac{d}{du}\left[\sec^{n-2}u\right]\cdot \tan u\ du$

$\displaystyle\longrightarrow I=\sec^{n-2}u\tan u-(n-2)\int\sec^{n-2}u\tan^2 u\ du$

$\displaystyle\longrightarrow I=\sec^{n-2}u\tan u-(n-2)\int\sec^{n-2}u\left(\sec^2u-1\right)\ du$

$\displaystyle\longrightarrow I=\sec^{n-2}u\tan u-(n-2)\int\left(\sec^nu-\sec^{n-2}u\right)\ du$

$\displaystyle\longrightarrow I=\sec^{n-2}u\tan u-(n-2)\int\left\sec^nu\ du+(n-2)\int\sec^{n-2}u\ du$

Replacing $\displaystyle\int\sec^nu\ du$ by $I,$

$\displaystyle\longrightarrow I=\sec^{n-2}u\tan u-(n-2)I+(n-2)\int\sec^{n-2}u\ du$

$\displaystyle\longrightarrow I+(n-2)I=\sec^{n-2}u\tan u+(n-2)\int\sec^{n-2}u\ du$

$\displaystyle\longrightarrow(n-1)I=\sec^{n-2}u\tan u+(n-2)\int\sec^{n-2}u\ du$

$\displaystyle\longrightarrow I=\dfrac{\sec^{n-2}u\tan u}{n-1}+\dfrac{n-2}{n-1}\int\sec^{n-2}u\ du$

That is,

$\displaystyle\longrightarrow\boxed{\int\sec^nu\ du=\dfrac{\sec^{n-2}u\tan u}{n-1}+\dfrac{n-2}{n-1}\int\sec^{n-2}u\ du,\quad\!n\neq1}$

Hence Verified!

Answer 2

Answer:

yeh hai

pinterst I'd meri

$f(x)=2x^3+6x^2+4x$

$g(x)=x^2+3x+2$

The polynomials f(x) and g(x) are defined above.

Which of the following polynomials is divisible by $2x+3$?

$A)h(x) = f(x) + g(x)$

$B) p(x) = f(x) + 3g(x)$

$C) r(x) = 2f(x) + 3g(x)$

$D) s(x) = 3f(x) + 2g(x)$