Question

Intrigation of root x sin x dx

Answer 1

Answer:

ıllıllı ʜᴇʏ ıllıllı

Step-by-step explanation:

Given ∫

√

sinx  dx

Let sinx = ⇔ $T^{2}$ cosx dx = 2tdt ⇔ dx =

$\frac{2T}\sqrt{1} - {T4}$dt

 

So Integral is 2∫$\frac{2T}\sqrt{1} - {t4}-\frac{1}{2}$dt

Now Using ∙∫xm.(a+bxn)pdx

where m,n,p are Rational no.

which is Integrable only when (

m+1

n

 

)∈Z or {

m+1

n

 

+p}∈Z

Now here 2∫t2.(1−t4)−

1

2

 

dt

m=2,a=1,b=−1,n=4,p=−

1

2

 

and (

2+1

4

 

)≠Z or (

2+1

4

 

)−

1

2

 

≠Z

So We can not integrate ∫

√

sinx

dx=2∫t2.(1−t4)−

1

2

 

dt in terms of elementry function.

(x) ^ 1/2 = t

now differentiating with respect to t we get,

dt/dx = 1/ 2(x)^1/2

now substituting this

we get, anti derivative of s sin( t) * 2 t dt

now. we will use integration by parts

which will give

2 [ t anti- deriv sin(t) - anti - deriv { d(t)/ dt anti- deriv sin(t) dt}dt

2 [ t * (-cost ) + anti- deriv cos(t) dt ]

2 [ t* (-cost) + sint dt ]

2 [ sint - tcost ]

now substituting the value of t , we get

Answer 2

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step-by-step explanation:

✍️ Integration is the reverse of differentiation.

However:

♦If y = 2x + 3,

dy/dx = 2

♦If y = 2x + 5,

dy/dx = 2

♦If y = 2x,

dy/dx = 2

So,

the integral of 2 can be 2x + 3, 2x + 5, 2x, etc.

✍️ For this reason, when we integrate, we have to add a constant.

✍️ So the integral of 2 is 2x + c, where c is a constant.

✍️ A "S" shaped symbol is used to mean the integral of, and dx is written at the end of the terms to be integrated, meaning "with respect to x".

✍️ This is the same "dx" that appears in dy/dx .

✍️ Now, The above question is solved in the attachment.

kindly refer to it.