Question

Evaluate integral 1/4-5sin^2x is multiply with dx
​

Answer 1

Step-by-step explanation:

We have,

$\int \frac{1}{4 - 5 \sin^{2} (x) }dx$

$= \int \frac{ \sec^{2} (x) }{4 \sec^{2} (x) - 5 \tan^{2} (x) } dx$

$= \int \frac{ \sec^{2} (x) }{4 + 4 \tan^{2} (x) - 5 \tan^{2} (x) } dx$

$= \int \frac{ \sec^{2} (x) }{4 - \tan^{2} (x) } dx$

Let, tan(x) = t

=> sec²(x) dx = dt

$= \int \frac{dt}{4 - {t}^{2} }$

$= \int \frac{dt}{ {(2)}^{2} - {(t)}^{2} }$

$= \frac{1}{2 \times 2} ln | \frac{2 + t}{2 - t} | + c$

$= \frac{1}{4} ln|\frac{2 - \tan(x) }{2 + \tan(x) } | + c$

$= ln| \sqrt[4]{ \frac{2 - \tan(x) }{2 + \tan(x) }} | + c$

Answer 2

Answer:

We have,

\begin{gathered}\int \frac{1}{4 - 5 \sin^{2} (x) }dx \\\end{gathered}

∫

4−5sin

2

(x)

1

dx

\begin{gathered}= \int \frac{ \sec^{2} (x) }{4 \sec^{2} (x) - 5 \tan^{2} (x) } dx\\\end{gathered}

=∫

4sec

2

(x)−5tan

2

(x)

sec

2

(x)

dx

\begin{gathered}= \int \frac{ \sec^{2} (x) }{4 + 4 \tan^{2} (x) - 5 \tan^{2} (x) } dx \\\end{gathered}

=∫

4+4tan

2

(x)−5tan

2

(x)

sec

2

(x)

dx

\begin{gathered}= \int \frac{ \sec^{2} (x) }{4 - \tan^{2} (x) } dx \\\end{gathered}

=∫

4−tan

2

(x)

sec

2

(x)

dx

Let, tan(x) = t

=> sec²(x) dx = dt

\begin{gathered}= \int \frac{dt}{4 - {t}^{2} } \\\end{gathered}

=∫

4−t

2

dt

\begin{gathered}= \int \frac{dt}{ {(2)}^{2} - {(t)}^{2} } \\\end{gathered}

=∫

(2)

2

−(t)

2

dt

\begin{gathered}= \frac{1}{2 \times 2} ln | \frac{2 + t}{2 - t} | + c \\\end{gathered}

=

2×2

1

ln∣

2−t

2+t

∣+c

\begin{gathered}= \frac{1}{4} ln|\frac{2 - \tan(x) }{2 + \tan(x) } | + c \\\end{gathered}

=

4

1

ln∣

2+tan(x)

2−tan(x)

∣+c

\begin{gathered}= ln| \sqrt[4]{ \frac{2 - \tan(x) }{2 + \tan(x) }} | + c \\\end{gathered}