Calculus Problem: integration of Cos[x^(1/3)] dx

[Seimuna]

im facing difficulty in solving

1) integration of Cos[x^(1/3)] dx...
i noe it is using integration by part, but i jus couldnt get the answer...
i wonder is step go wrong or the answer given is wrong...
this is the answer given -> 6[x^(1/3)] Cos[x^(1/3)] + 3(x^(2/3)) Sin[x^(1/3)] - 6 Sin[x^(1/3)]...
my step start with u=Sin[x^(1/3)] and dv=dx...

2) integration of (x^2) Sqrt[x^2 + 9] dx
i manage get integration of (sec^5 tita - sec^3 tita d(tita))
and i dono how to continue do this...

hope some1 can help me...
Thanks in advance...

 

[Deleted member 4993]

Re: integration of Cos[x^(1/3)] dx

im facing difficulty in solving integration of Cos[x^(1/3)] dx...
i noe it is using integration by part, but i jus couldnt get the answer...
i wonder is step go wrong or the answer given is wrong...
this is the answer given -> 6[x^(1/3)] Cos[x^(1/3)] + 3(x^(2/3)) Sin[x^(1/3)] - 6 Sin[x^(1/3)]...
my step start with u=Sin[x^(1/3)] and dv=dx...

Thanks in advance...

\(\displaystyle \int\cos(x^{\frac{1}{3}})dx\)

sustitute

u = x^(1/3)

du = (1/3)x(-2/3)dx

dx = 3u^2du

so

\(\displaystyle \int\cos(x^{\frac{1}{3}})dx\)

\(\displaystyle = \, 3\int u^2\cdot\cos(u)du\)

Now do integration by parts and substitute back 'x'.

 

[Deleted member 4993]

Re: Calculus Problem

im facing difficulty in solving

1) integration of Cos[x^(1/3)] dx...
i noe it is using integration by part, but i jus couldnt get the answer...
i wonder is step go wrong or the answer given is wrong...
this is the answer given -> 6[x^(1/3)] Cos[x^(1/3)] + 3(x^(2/3)) Sin[x^(1/3)] - 6 Sin[x^(1/3)]...
my step start with u=Sin[x^(1/3)] and dv=dx...

2) integration of (x^2) Sqrt[x^2 + 9] dx
i manage get integration of (sec^5 tita - sec^3 tita d(tita)) <<< Please show work
and i dono how to continue do this...

hope some1 can help me...
Thanks in advance...

 

[Seimuna]

Re: Calculus Problem

integration of (x^2) (Sqrt [x^2 + 9]) dx

erm...
let x = 3 tan tita for tan tita = x/3
dx = 3 sec^2 tita d(tita)

substitute in, then ll get
81 integration of (tan^2 tita) (3 sec tita) (3 sec^2 tita d(tita))
then substitute tan^2 tita = sec^2 tita -1

then in the end ll get 81 integration of (sec^5 tita + sec^3 tita)
then i dono how to continue... =p

 

[Seimuna]

Re: integration of Cos[x^(1/3)] dx

\(\displaystyle \int\cos(x^{\frac{1}{3}})dx\)

sustitute

u = x^(1/3)

du = (1/3)x(-2/3)dx

dx = 3u^2du

so

\(\displaystyle \int\cos(x^{\frac{1}{3}})dx\)

\(\displaystyle = \, 3\int u^2\cdot\cos(u)du\)

Now do integration by parts and substitute back 'x'.

thanks...i got the answer...^^

 

[Deleted member 4993]

Re: Calculus Problem

integration of (x^2) (Sqrt [x^2 + 9]) dx

erm...
let x = 3 tan tita for tan tita = x/3
dx = 3 sec^2 tita d(tita)

substitute in, then ll get
81 integration of (tan^2 tita) (3 sec tita) (3 sec^2 tita d(tita))
then substitute tan^2 tita = sec^2 tita -1

then in the end ll get 81 integration of (sec^5 tita + sec^3 tita)
then i dono how to continue... =p

\(\displaystyle \int x^2\cdot\sqrt{x^2+9}dx\)

\(\displaystyle = \frac{1}{3}\int x\cdot 3x\cdot\sqrt{x^2+9}dx\)

\(\displaystyle = \, x\cdot (x^2+9)^{\frac{3}{2}} - \int (x^2+9)^{\frac{3}{2}} dx\)

Now do the substitution...

 

[Deleted member 4993]

Re: Calculus Problem

\(\displaystyle \int \sec^3(x) dx=\tan(x).sec(x) -\int tan^2(x).sec(x)dx=tan(x).sec(x) - \int[sec^2(x)-1].sec(x)dx\)

\(\displaystyle \int \sec^3(x) dx \, = \, \frac{1}{2}[tan(x).sec(x) + ln[|sec(x)+tan(x)|]]\)

similarly you can calculate sec^5(x)

The formula is:

\(\displaystyle \int \sec^n(x) dx \, = \, \frac{sec^{n-2}(x)\cdot tan(x)}{n-1} + \frac{n-2}{n-1}\int sec^{n-2}(x) dx ....for \, n \ge 2\)

 

[Seimuna]

Re: Calculus Problem

\(\displaystyle \int x^2\cdot\sqrt{x^2+9}dx\)

\(\displaystyle = \frac{1}{3}\int x\cdot 3x\cdot\sqrt{x^2+9}dx\)

\(\displaystyle = \, x\cdot (x^2+9)^{\frac{3}{2}} - \int (x^2+9)^{\frac{3}{2}} dx\)

Now do the substitution...

im so sorry...
erm...i don und y u need to separate the (x)(3x) in the second step...
tat makes me cant solve it into the third step...

 

[Seimuna]

Re: Calculus Problem

\(\displaystyle \int \sec^3(x) dx=\tan(x).sec(x) -\int tan^2(x).sec(x)dx=tan(x).sec(x) - \int[sec^2(x)-1].sec(x)dx\)

\(\displaystyle \int \sec^3(x) dx \, = \, \frac{1}{2}[tan(x).sec(x) + ln[|sec(x)+tan(x)|]]\)

similarly you can calculate sec^5(x)

The formula is:

\(\displaystyle \int \sec^n(x) dx \, = \, \frac{sec^{n-2}(x)\cdot tan(x)}{n-1} + \frac{n-2}{n-1}\int sec^{n-2}(x) dx ....for \, n \ge 2\)

y in ur formula, the second integration didnt include tan ?

 

[Deleted member 4993]

Re: Calculus Problem

\(\displaystyle \int \sec^3(x) dx=\tan(x).sec(x) -\int tan^2(x).sec(x)dx=tan(x).sec(x) - \int[sec^2(x)-1].sec(x)dx\)

\(\displaystyle \int \sec^3(x) dx \, = \, \frac{1}{2}[tan(x).sec(x) + ln[|sec(x)+tan(x)|]]\)

similarly you can calculate sec^5(x)

The formula is:

\(\displaystyle \int \sec^n(x) dx \, = \, \frac{sec^{n-2}(x)\cdot tan(x)}{n-1} + \frac{n-2}{n-1}\int sec^{n-2}(x) dx ....for \, n \ge 2\)

y in ur formula, the second integration didnt include tan ?

Correct - it has only sec(x) - similar to the function you started from.

I cannot tell you "Y" - but the simplest answer is - that 's the way it is.

 

[Seimuna]

\(\displaystyle \int x^2\cdot\sqrt{x^2+9}dx\)

\(\displaystyle = \frac{1}{3}\int x\cdot 3x\cdot\sqrt{x^2+9}dx\)

\(\displaystyle = \, x\cdot (x^2+9)^{\frac{3}{2}} - \int (x^2+9)^{\frac{3}{2}} dx\)

Now do the substitution...

im so sorry...
erm...i don und y u need to separate the (x)(3x) in the second step...
tat makes me cant solve it into the third step...