Use the trigonometric substitution u = a tan θ,where 0

Question

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

Quizplus · Accepted Answer

We can start by using the trigonometric identity:

$\sec^{2}	heta = 	an^{2}	heta + 1$

Multiplying both sides by $a^{2}$, we get:

$a^{2}\sec^{2}	heta = a^{2}	an^{2}	heta + a^{2}$

Now, we can use the given substitution $u = a	an	heta$, which means $	an	heta = \frac{u}{a}$. Substituting this into the equation above, we get:

$a^{2}\sec^{2}	heta = u^{2} + a^{2}$

Solving for $\sec	heta$, we get:

$\sec	heta = \frac{\sqrt{u^{2} + a^{2}}}{a}$

Therefore, our simplified expression is:

$\int\frac{11e^{a^{2}	an^{2}	heta}}{a}d	heta = \int\frac{11e^{u^{2}}}{a\sqrt{u^{2} + a^{2}}}du$

And substituting back $	heta = 	an^{-1}\left(\frac{u}{a}ight)$:

$\int\frac{11e^{a^{2}	an^{2}	heta}}{a}d	heta = \int\frac{11e^{u^{2}}}{a\sqrt{u^{2} + a^{2}}}du = \int\frac{11e^{	an^{2}	heta}}{a\sec	heta}d\left(	an^{-1}\frac{u}{a}ight) = \int\frac{11e^{	an^{2}	heta}}{a}\cos	heta d	heta$

Recall that we have the identity $\sec^{2}	heta = 	an^{2}	heta + 1$, so $	an^{2}	heta = \sec^{2}	heta - 1$. Substituting this into our expression, we get:

$\int\frac{11e^{	an^{2}	heta}}{a}\cos	heta d	heta = \int\frac{11e^{\sec^{2}	heta - 1}}{a}\cos	heta d	heta = \int\frac{11e^{u^{2} - a^{2}}}{a\sqrt{u^{2} + a^{2}}}du$

This integral can be evaluated using the substitution $t = u^{2} + a^{2}$, which gives:

$\int\frac{11e^{u^{2} - a^{2}}}{a\sqrt{u^{2} + a^{2}}}du = \frac{11}{a}\int\frac{e^{t}}{\sqrt{t}}dt = 11a	ext{Ei}\left(\frac{t}{2}ight) + C$

Substituting back $t = u^{2} + a^{2}$, we get:

$\int\frac{11e^{	an^{2}	heta}}{a}\cos	heta d	heta = 11a	ext{Ei}\left(\frac{u^{2} + a^{2}}{2}ight) + C$

Therefore, the answer is (B) $a\sec	heta$, which is equivalent to $\sqrt{u^{2} + a^{2}}$.

Use the Trigonometric Substitution U = a Tan θ,Where 0 a2+u2\sqrt { a ^ { 2 } + u ^ { 2 } }a2+u2​

Use the Trigonometric Substitution U = a Tan θ,Where 0 $\sqrt { a ^ { 2 } + u ^ { 2 } }$