CK-12-Calculus

6.5. Derivatives and Integrals Involving Inverse Trigonometric Functions http://www.ck12.org

To change the limits,

x=ln 2→u=e−x=e−ln 2=eln 1/^2 =^12 ,

x=ln

( 2

√ 3

)

→u=e−x=e−ln 2/

√ 3

=

√ 3

2.

Thus our integral becomes

∫ln( 2 /√ 3 )

ln2

√e−x

1 −e−^2 xdx=

∫ √ 3 / 2

1 / 2

√u

1 −u^2

−du

u

=−

∫√ 3 / 2

1 / 2

√^1

1 −u^2 du

=−

[

sin−^1 u

]√ 3 / 2

1 / 2

=−

[

sin−^1

(√

3

2

)

−sin−^1

( 1

2

)]

=−

[π

3 −

π

6

]

=−π 6.

Multimedia Links

For a video presentation of the derivatives of inverse trigonometric functions(4.4), see Math Video Tutorials by
James Sousa, The Derivatives of Inverse Trigonometric Functions (8:55).

MEDIA

Click image to the left for use the URL below.

URL: http://www.ck12.org/flx/render/embeddedobject/589

For three presentations of integration involving inverse trigonometric functions(18.0), see Math Video Tutorials by
James Sousa, Integration Involving Inverse Trigonometric Functions, Part1 (7:39)

MEDIA

Click image to the left for use the URL below.

URL: http://www.ck12.org/flx/render/embeddedobject/590