Use the trigonometric substitution u = a sin θ,where - π/2

Question

Use the trigonometric substitution u = a sin θ,where - π/2 < θ < π/2 and a > 0 to simplify the expression $\sqrt { a ^ { 2 } - u ^ { 2 } }$ . ​
A)​a csc θ
B)​a sin θ
C)​a tan θ
D)​a cos θ
E)​- a cos θ

Quizplus · Accepted Answer

Using the trigonometric substitution u = a sin θ, we can find that:
- du/dθ = a cos θ
- sin θ = u/a
- cos θ = √(1- sin^2 θ) = √(1- u^2/a^2)

Substituting these expressions into the given expression, we get:

∫ 11ea99db_b4a0_dc65_94ac_b9172d5edfd9_TB4590_11 dx
= ∫ a^2 cos^2 θ / √(a^2 - a^2 sin^2 θ) dx
= ∫ a^2 cos^2 θ / √(a^2 - u^2) (du/a)
= ∫ a cos^2 θ du/ √(a^2-u^2)

Now using the identity cos^2 θ = 1 - sin^2 θ, we get:

∫ a cos^2 θ du/ √(a^2-u^2) 
= ∫ a (1 - sin^2 θ) du/ √(a^2-u^2) 
= ∫ [a/√(a^2-u^2) - a sin^2 θ /√(a^2-u^2)] du

Substituting back the expressions for sin θ and cos θ, we get:

∫ [1/√(1-u^2/a^2) - u^2/√(a^2-u^2)] du

This is the integral of the standard form for the derivative of arccos(u/a). Therefore, the answer is:

∫ 11ea99db_b4a0_dc65_94ac_b9172d5edfd9_TB4590_11 dx = -a arccos(u/a) + C

Substituting back u = a sin θ, we get:

∫ 11ea99db_b4a0_dc65_94ac_b9172d5edfd9_TB4590_11 dx = -a arccos(sin θ) + C = -a cos θ + C

Therefore, the best choice is D) a cos θ.

Use the Trigonometric Substitution U = a Sin θ,Where - a2−u2\sqrt { a ^ { 2 } - u ^ { 2 } }a2−u2​

Use the Trigonometric Substitution U = a Sin θ,Where - $\sqrt { a ^ { 2 } - u ^ { 2 } }$