Answer:
(a) ⅛ tan⁻¹(¼)
(b) sec x − ln│csc x + cot x│+ C
Step-by-step explanation:
(a) ∫₀¹ x / (16 + x⁴) dx
∫₀¹ (x/16) / (1 + (x⁴/16)) dx
⅛ ∫₀¹ (x/2) / (1 + (x²/4)²) dx
If tan u = x²/4, then sec²u du = x/2 dx
⅛ ∫ sec²u / (1 + tan²u) du
⅛ ∫ du
⅛ u + C
⅛ tan⁻¹(x²/4) + C
Evaluate from x=0 to x=1.
⅛ tan⁻¹(1²/4) − ⅛ tan⁻¹(0²/4)
⅛ tan⁻¹(¼)
(b) ∫ (sec³x / tan x) dx
Multiply by cos x / cos x.
∫ (sec²x / sin x) dx
Pythagorean identity.
∫ ((tan²x + 1) / sin x) dx
Divide.
∫ (tan x sec x + csc x) dx
Split the integral
∫ tan x sec x dx + ∫ csc x dx
Multiply second integral by (csc x + cot x) / (csc x + cot x).
∫ tan x sec x dx + ∫ csc x (csc x + cot x) / (csc x + cot x) dx
Integrate.
sec x − ln│csc x + cot x│+ C
Answer:
(a) Solution : 1/8 cot⁻¹(4) or 1/8 tan⁻¹(¼) (either works)
(b) Solution : tan(x)/sin(x) + In | tan(x/2) | + C
Step-by-step explanation:
(a) We have the integral (x/16 + x⁴)dx on the interval [0 to 1].
For the integrand x/6 + x⁴, simply pose u = x², and du = 2xdx, and substitute:
1/2 ∫ (1/u² + 16)du
'Now pose u as 4v, and substitute though integral substitution. First remember that we have to factor 16 from the denominator, to get 1/2 ∫ 1/(16(u²/16 + 1))' :
∫ 1/4(v² + 1)dv
'Use the common integral ∫ (1/v² + 1)dv = arctan(v), and substitute back v = u/4 to get our solution' :
1/4arctan(u/4) + C
=> Solution : 1/8 cot⁻¹(4) or 1/8 tan⁻¹(¼)
(b) We have the integral ∫ sec³(x)/tan(x)dx, which we are asked to evaluate. Let's start by substitution tan(x) as sin(x)/cos(x), if you remember this property. And sec(x) = 1/cos(x) :
∫ (1/cos(x))³/(sin(x)/cos(x))dx
If we cancel out certain parts we receive the simplified expression:
∫ 1/cos²(x)sin(x)dx
Remember that sec(x) = 1/cos(x):
∫ sec²(x)/sin(x)dx
Now let's start out integration. It would be as follows:
[tex]\mathrm{Let:u=\frac{1}{\sin \left(x\right)},\:v'=\sec ^2\left(x\right)}\\=> \frac{\tan \left(x\right)}{\sin \left(x\right)}-\int \:-\cot \left(x\right)\csc \left(x\right)\tan \left(x\right)dx\\\\\int \:-\cot \left(x\right)\csc \left(x\right)\tan \left(x\right)dx=-\ln \left|\tan \left(\frac{x}{2}\right)\right|\\=> \frac{\tan \left(x\right)}{\sin \left(x\right)}-\left(-\ln \left|\tan \left(\frac{x}{2}\right)\right|\right)\\[/tex]
[tex]=> \frac{\tan \left(x\right)}{\sin \left(x\right)}+\ln \left|\tan \left(\frac{x}{2}\right)\right|\\\\=> \frac{\tan \left(x\right)}{\sin \left(x\right)}+\ln \left|\tan \left(\frac{x}{2}\right)\right|+C[/tex]
Solution: tan(x)/sin(x) + In | tan(x/2) | + C
